linear equations difficult problems

Linear Equations Word Problems Worksheet with Solutions

The first equation family that many students learn about is linear equations. Linear equations have many applications in the real-world, which can make for a really great set of word problems! 

As a student studying algebra, you will encounter many linear equations word problems. That’s why I have put together this linear equations word problems worksheet with solutions!

My hope is that this linear equation word problems worksheet and answer key help you deepen your understanding of linear equations and linear systems!

What Are Linear Equations?

A linear equation is an  algebraic equation  where the highest power on the variable is one. When graphed, a linear equation will produce a straight line.

There are a few ways that we can write linear equations, with two of the most common being slope-intercept form and standard form .

Slope-intercept form is best way to identify the slope of the line and the y-intercept of the line. In general, the equation of a line in slope-intercept form is written as:

In this form,  a  represents the slope of the line and  b  represents the y-intercept of the line.

Equations of lines in standard form are easy to recognize because it is a uniformly recognized form of a line. Standard form allows for easy comparison of coefficients. When two linear equations are in standard form, you can quickly compare the coefficients of x and y.

In general, the standard form of a line is written as:

$$Ax+By=C$$

Note that A and B do not represent the slope of the line or the y-intercept in this form. Instead, A and B are simply constants.

Solving Linear Equations

Any set of word problems relating to linear equations will ask you to solve an equation of some sort. However, there are many different types of solving equations problems that you will encounter as you explore linear equations word problems.

Let’s take a look at a few different types to make sure you know what to expect when you check out the linear equation word problems worksheet with solutions below.

Solving Two-Step Equations and Multi-Step Equations

One of the simplest equation problems that you can solve is a two-step equation. A two-step equation requires you to perform just two steps in order to determine the unique solution to the linear equation.

The first step is to identify the side of the equation with the unknown variable. Your goal will be to isolate this variable (or get it by itself). Consider the following example:

In this example, the variable is on the right-hand side of the equation. To isolate x, we “undo” the operations on the right-hand side of the equation using inverse operations. This just means doing whatever the opposite operation is.

We can apply  order of operations  in reverse to start with the subtraction and then deal with the multiplication. Adding 5 to both sides and then dividing by 2 will result in:

$$\begin{split} 9+5&=2x-5+5 \\ \\ 14 &= 2x \\ \\ \frac{14}{2} &= \frac{2x}{2} \\ \\ 7 &= x \end{split} $$

This shows that the value of the unknown variable here is x = 7. This is a unique solution that will satisfy the equation.

If you want to learn more about finding the solution of linear equations and explore multi-step equations that use the distributive property, check out  these equation solving worksheets !

Systems of Linear Equations

Systems of linear equations are another type of problem that you will see on the linear equations word problems worksheet linked below. A system of linear equations involves two (or more!) linear equations that intersect in some way. 

There are a few different ways that linear equations can intersect :

• Once  at a single point of intersection
• Never  as a result of the lines being parallel 
• Always  as a result of the lines being on top of one another

When solving a system of linear equations, your goal is to determine both unknown variables. If the lines intersect, the solution to the system will be the point of intersection for the lines.

When given a linear equation, we can find the point of intersection between it and a second equation using a few different methods. I made a  video on the substitution method  and a  video on the elimination method  to help you understand these strategies for solving systems before you apply them to word problems involving systems of equations.

What Are Linear Equation Word Problems?

A linear equation word problem involves a real-world situation or scenario that can be solved by setting up and solving linear equations. The equations that are used model the relationships between different quantities in the real-world scenario. 

The topics of these problems vary, ranging from applications in science and physics (ie. calculate the speed of the boat) to business applications (ie. how many sales are required to break even?). The problems that you encounter will also vary in depth and difficulty. 

In my teaching experience, students tend to struggle with word problems because it isn’t always immediately clear what is being asked.  I have seen many students feel very confident in their equation solving skills, yet they still struggle when it comes to solving linear equation word problems.

One reason for this is that you aren’t always given equations from the start while solving word problems.

Tips for Solving Linear Equation Word Problems

During my time as a high school math teacher, I have come across a few tips that I think will help you solve linear equation word problems successfully.

To begin, the first step should always be to define two variables. Read the question carefully and think about the quantities involved. Use variables to represent these quantities.

The second step should be writing an equation that models the scenario. Depending on the problem, you may need to write a second equation as well.

Lastly, think about what the problem is asking you to find. 

For example, are you looking for the values of two unknown variables? If so, you are likely going to be setting up and solving a system of linear equations.

​If you are being asked for the value of a single variable, the chances are you will be solving a single linear equation.

Now that you have a basic understanding of the concepts involved in solving linear equations word problems, it’s time to try a few!

My goal here is to provide you with a worksheet that you can use to practice and feel confident that you understand linear equations word problems!

While I was writing this worksheet, I made sure to include a wide variety of problems that range in difficulty. You will see a few simpler problems involving a two-step equation or multi-step equations, but you will also see a few problems that involve systems of linear equations.

After solving each word problem, be sure to check the answer key to verify that you fully understand the process used to set up the problem and solve it. Reflecting on your understanding is an important part of developing comfort with any given math concept!

Click below to download the linear equations word problems worksheet with solutions!

Using This Linear Equation Word Problems Worksheet

Being able to read a  real-world algebra problem  and set up a linear equation (or a system of linear equations) to solve it is a very challenging skill. In my experience as a math teacher, many students struggle with this concept, even if they fully understand the mathematics that the problem requires.

This is the main reason that I put together this linear equation word problems worksheet with solutions. My goal is to provide you with a set of word problems that you can use to check your understanding of solving linear equations in the real-world.

I hope you found this practice worksheet helpful as you continue your studies of algebra and linear equations!

If you are looking for more linear equations math worksheets in PDF formats, check out my collection of  solving linear inequalities worksheets  and this  linear inequality word problems worksheet .

Did you find this linear equation word problems worksheet with solutions helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!

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Examples of One Solution Equations, Zero, and Infinite

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Course: Algebra 1   >   Unit 2

• Why we do the same thing to both sides: Variable on both sides
• Intro to equations with variables on both sides
• Equations with variables on both sides: 20-7x=6x-6

Equations with variables on both sides

• Equation with variables on both sides: fractions
• Equations with variables on both sides: decimals & fractions
• Equation with the variable in the denominator

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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Linear Equations in Two Variables Questions

Linear equations in two variables questions presented here cover a variety of questions asked regarding linear equations in two variables with solutions and proper explanations. By practising these questions students will develop problem-solving skills.

Linear equations in two variables are linear polynomials with two unknowns. They are of the general form ax + by + c = 0, where x and y are the two variables, a and b are non-zero real numbers and c is a constant. The graphical representation of a linear equation in two variables is a straight line.

Linear Equations in Two Variables Questions with Solutions

Below are some practice questions on linear equations in two variables with detailed solutions.

Question 1: Solve for x and y:

\(\begin{array}{l}\frac{1}{2x}-\frac{1}{y}=-1,\:\:\frac{1}{x}+\frac{1}{2y}=8\:\:\:(x\neq0,\;y\neq0)\end{array} \)

Put 1/x = u and 1/y = v. The given equations become

u/2 – v = –1 ⇒ u – 2v = – 2 ….(i)

u + v/2 = 8 ⇒ 2u + v = 16 ….(ii)

Multiplying equation (ii) by 2 on both sides and adding (i) and (ii), we get

(u + 4u) + ( –2v + 2v) = –2 + 32

⇒ u = 6 ⇒ x = ⅙ and y = ¼

Question 2: Solve the system of linear equations 2x + 3y = 17 and 3x – 2y = 6 by the cross multiplication method.

By cross multiplication

\(\begin{array}{l}\frac{x}{\left\{ 3\times (-6)-(-2)\times(-17)\right\}}=\frac{y}{\left\{ -17\times 3-(-6)\times 2 \right\}}=\frac{1}{\left\{2\times (-2)-3 \times3\right\}}\end{array} \)

⇒ x/( –52) = y/( –39) = 1/( – 13)

⇒ x = 52/13 = 4 and y = 39/13 = 3

Hence x = 4 and y = 3 is the solution of given equations.

Question 3: Solve the following system of equations by substitution method:

2x + 3y = 0 and 3x + 4y = 5

Given equations,

2x + 3y = 0 ….(i)

3x + 4y = 5 …..(ii)

From (i) we get, y = – 2x/3, substituting value of y in (ii), we get

3x + 4(–2x/3) = 5

⇒ 9x – 8x = 15

Then y = (–2 × 15)/3 = – 10

Therefore, x = 15 and y = – 10 is solution of given system of equations.

Video Lesson on Consistent and Inconsistent Equations

Question 4: Find the value of k for which the given system of equations has infinitely many solutions: x + (k + 1)y = 5 and (k + 1)x + 9y + (1 – 8k) = 0.

The given equations will have infinitely many solutions if a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Hence, 1/(k + 1) = (k + 1)/9 = – 5/(1 – 8k)

Solving the equations we get k = 2.

Question 5: If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel. Find the value of k.

If the lines are parallel, then they are inconsistent system of equations and a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2

Now, it should be 3/2 = 2k/5

Then we have 2k/5 = 30/20 = 3/2, which satisfies the condition of inconsistency.

Question 6: Find the value of k for which the system of equations has a non-zero solution

5x + 3y = 0 and 10x + ky = 0

The given equations are homogenous equations, they will have a non-zero solution if a 1 /a 2 = b 1 /b 2

Then, 5/10 = 3/k

⇒ 1/2 = 3/k

Question7: The monthly incomes of A and B are in the ratio 8:7 and their expenditures are in the ratio 19:16. If each saves ₹ 5000 per month, find the monthly income of each.

Let the monthly incomes of A and B be 8x and 7x rupees respectively, and let their monthly expenditure be 19y and 16y rupees respectively.

Monthly savings of A = 8x – 19y = 5000 ….(i)

Monthly savings of B = 7x – 16y = 5000 ….(ii)

Multiplying (i) 16 and (ii) by 19 and subtracting (ii) from (i) we get

(16 × 8x – 19 × 7x) = 5000 (16 – 19)

⇒ 5x = 15000 ⇒ x = 3000

Monthly income of A is (8 × 3000) = ₹24,000

Monthly income of B is (7 × 3000) = ₹21,000

Question 8: The sum of a two-digit number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the original number.

Let the original number be (10x + y)

According to the question,

(10x + y) + (10y + x) = 99

⇒ 11(x + y) = 99

⇒ x + y = 9 ….(i)

And x – y = 3 ….(ii)

Adding equations (i) and (ii), we get,

Hence the required number is 63.

Question 9: A man’s age is three times the sum of the ages of his two sons. After 5 years, his age will be twice the sum of his two son’s age. Find the age of the man.

Let the age of the man be x and the sum of the ages of his two sons be y.

x = 3y ⇒ x – 3y = 0 ….(i)

And (x + 5) = 2(y + 5 + 5)

⇒ x – 2y = 15 ….(ii)

Subtracting equation (i) from (ii) we get

Y = 15 and from (i) x = 45.

The present age of the man is 45 years.

Question 10: A man can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours, Find his speed of rowing in still water. Also, find the speed of the stream.

Let the speed of the man in still water be x km/hr and let the speed of the current be y kn/hr.

Speed in downstream = (x + y) km/hr

Speed in upstream = (x – y) km/hr

But speed in downstream = 20/2 km/hr = 10 km/hr

And speed in upstream = 4/2 km/hr = 2 km/hr

∴ x + y = 10 and x – y = 2

Solving both the equations we get;

x = 6 and y = 4.

Hence, the speed of the man in still water is 6 km/hr and the speed of the current is 4 km/hr.

Question 11: Find the four angles of a cyclic Quadrilateral ABCD in which ∠A = (2x – 1) o , ∠B = (y + 5) o , ∠C = (2y + 15) o , and ∠D = (4x – 7) o .

We know that sum of opposite angles of a cyclic quadrilateral is 180 o

∴ ∠A + ∠C = 180 o and ∠B + ∠C = 180 o

∠A + ∠C = 180 o ⇒ (2x – 1) + (2y + 15) = 180 o

⇒ x + y = 83 ….(i)

∠B + ∠C = 180 o ⇒ (y + 5) + (4x – 7) = 180 o

⇒ 4x + y = 182 …..(ii)

Subtracting (i) from (ii) we get

3x = 182 – 83 ⇒ x = 33

Substituting in (i), we get y = 50

∴ ∠A = 2 × 33 – 1 = 65 o

∠B = 50 + 5 = 55 o

∠C = 2 × 50 + 15 = 115 o

∠D = 4 × 33 – 7 = 125 o .

Question 12: 8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find time taken by a man and a boy alone to finish the same work.

Let 1 man can finish the work in x days and let 1 boy can finish the work in y days.

1 man’s one day work = 1/x

1 boy’s one day work = 1/y

8 men’s 1 day’s work + 12 boy’s one day’s work = ⅕

⇒ 8/x + 12/y = ⅕

⇒ 8u + 12v = ⅕ ….(i) where u = 1/x and v = 1/y

Similarly, 6u + 8v = 1/7 ….(ii)

On solving (i) and (ii) we get, x = 70 and y = 140

∴ One man alone can finish the work in 70 days and one boy alone can finish the work in 140 days.

Related Articles:

Practice questions:.

1. Five years ago Anna was three times older than Mira and ten years later Anna will be two times older than Mira. What are the present ages of Anna and Mira?

2. The difference of two numbers is 4 and the difference of their reciprocals is 4/21. Find the numbers.

3. Find the value of k for which the system of equations 5x – 3y = 0, and 2x + ky = 0 has a non-zero solution.

4. Find the value of a and b for which each of the following systems of linear equations

(a – 1)x + 3y = 2 and 6x + (1 – 2b)y = 6 has infinite number of solutions.

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1.3: Linear Equations in One Variable

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Learning Objectives

• Solve equations in one variable algebraically.
• Solve a rational equation.
• Find a linear equation.
• Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays \($15.00/hr\), and she opened a savings account with an initial deposit of \($400\) on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately \($2,500\), how many hours will she have to work to earn enough to pay for her vacation? If she can only work \(4\) hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure \(\PageIndex{1}\).

Solving Linear Equations in One Variable

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is \(1\). Linear equations in one variable may take the form \(ax +b=0\) and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent.

• An identity equation is true for all values of the variable. Here is an example of an identity equation: \[3x=2x+x \nonumber \] The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for \(x\) will make the equation true.
• A conditional equation is true for only some values of the variable. For example, if we are to solve the equation \(5x+2=3x−6\), we have the following: \[\begin{align*} 5x+2&=3x-6 \\ 2x &=-8 \\ x&=-4 \end{align*} \] The solution set consists of one number: \({−4}\). It is the only solution and, therefore, we have solved a conditional equation.
• An inconsistent equation results in a false statement. For example, if we are to solve \(5x−15=5(x−4)\), we have the following: \[\begin{align*} 5x−15 &=5x−20 \\ 5x−15-5x &= 5x−20-5x \\ −15 &\neq −20 \end{align*}\]Indeed, \(−15≠−20\). There is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.

LINEAR EQUATION IN ONE VARIABLE

A linear equation in one variable can be written in the form

where a and b are real numbers, \(a≠0\).

Howto: Given a linear equation in one variable, use algebra to solve it

The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads \(x=\)_________, if \(x\) is the unknown. There is no set order, as the steps used depend on what is given:

• We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
• Apply the distributive property as needed: \(a(b+c)=ab+ac\).
• Isolate the variable on one side of the equation.
• When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Example \(\PageIndex{1}\): Solving an Equation in One Variable

Solve the following equation: \(2x+7=19\).

This equation can be written in the form \(ax +b=0\) by subtracting 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

\[\begin{align*} 2x+7&=19\\ 2x&=12\qquad \text{Subtract 7 from both sides}\\ x&=6\qquad \text{Multiply both sides by } \dfrac{1}{2} \text{ or divide by } 2 \end{align*}\]

The solution is \(6\).

Exercise \(\PageIndex{1}\)

Solve the linear equation in one variable: \(2x+1=−9\).

\(x=−5\)

Example \(\PageIndex{2}\): Solving When the Variable Appears on Both Sides

Solve the following equation: \(4(x−3)+12=15−5(x+6)\).

Apply standard algebraic properties.

\[\begin{align*} 4(x-3)+12&=15-5(x+6)\\ 4x-12+12&=15-5x-30\qquad \text{Apply the distributive property}\\ 4x&=-15-5x\qquad \text{Combine like terms}\\ 9x&=-15\qquad \text{Place x terms on one side and simplify}\\ x&=-\dfrac{15}{9}\qquad \text{Multiply both sides by } \dfrac{1}{9} \text { , the reciprocal of } 9\\ x&=-\dfrac{5}{3} \end{align*}\]

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, \(x=-\dfrac{5}{3}\).

Exercise \(\PageIndex{2}\)

Solve the equation in one variable: \(−2(3x−1)+x=14−x\).

Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation . Recall that a rational number is the ratio of two numbers, such as \(\dfrac{2}{3}\) or \(\dfrac{7}{2}\). A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

\[\dfrac{x+1}{x^2-4} \nonumber \]

\[\dfrac{1}{x-3} \nonumber \]

\[\dfrac{4}{x^2+x-2} \nonumber \]

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Example \(\PageIndex{3}\): Solving a Rational Equation

Solve the rational equation:

\[\dfrac{7}{2x}-\dfrac{5}{3x}=\dfrac{22}{3} \nonumber \]

We have three denominators; \(2x\),\(3x\), and \(3\). The LCD must contain \(2x\),\(3x\), and \(3\). An LCD of \(6x\) contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD \(6x\).

\[\begin{align*} (6x)\left[\dfrac{7}{2x}-\dfrac{5}{3x}\right]&=\left[\dfrac{22}{3}\right](6x)\\ (6x)\left(\dfrac{7}{2x}\right)-(6x)\left(\dfrac{5}{3x}\right)&=\left(\dfrac{22}{3}\right)(6x)\qquad \text{Use the distributive property. Cancel out the common factors}\\ 3(7)-2(5)&=22(2x)\qquad \text{Multiply remaining factors by each numerator.}\\ 21-10&=44x\\ 11&=44x\\ \dfrac{11}{44}&=x\\ \dfrac{1}{4}&=x \end{align*}\]

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as \((x+1)\). Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are \(x\), \(x−1\), and \(3x−3\). First, factor all denominators. We then have \(x\), \((x−1)\), and \(3(x−1)\) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of \((x−1)\). The x in the first denominator is separate from the \(x\) in the \((x−1)\) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the \(x\), one factor of \((x−1)\), and the 3. Thus, the LCD is the following:

\(x(x−1)3=3x(x−1)\)

So, both sides of the equation would be multiplied by \(3x(x−1)\). Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as \(x\) and \(x^2+2x\). Once the second denominator is factored as \(x^2+2x=x(x+2)\), there is a common factor of \(x\) in both denominators and the LCD is \(x(x+2)\).

Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

\[\dfrac{a}{b}=\dfrac{c}{d}\]

We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.

Multiply a(d) and b(c), which results in \(ad=bc\).

Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

RATIONAL EQUATIONS

A r ational equation contains at least one rational expression where the variable appears in at least one of the denominators.

Howto: Given a rational equation, solve it.

• Factor all denominators in the equation.
• Find and exclude values that set each denominator equal to zero.
• Find the LCD.
• Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
• Solve the remaining equation.
• Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator

Example \(\PageIndex{4}\): Solving a Rational Equation without Factoring

Solve the following rational equation:

\(\dfrac{2}{x}-\dfrac{3}{2}=\dfrac{7}{2x}\)

We have three denominators: \(x\), \(2\), and \(2x\). No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is \(2x\). Only one value is excluded from a solution set, \(0\). Next, multiply the whole equation (both sides of the equal sign) by \(2x\).

\[\begin{align*} 2x\left[\dfrac{2}{x}-\dfrac{3}{2}\right]&=\left[\dfrac{7}{2x}\right](2x)\\ 2x\left(\dfrac{2}{x}\right)-2x\left(\dfrac{3}{2}\right)&=\left(\dfrac{7}{2x}\right)(2x)\qquad \text{Distribute } 2x\\ 2(2)-3x&=7\qquad \text{Denominators cancel out.}\\ 4-3x&=7\\ -3x&=3\\ x&=-1 \text { or } \{-1\} \end{align*}\]

The proposed solution is \(−1\), which is not an excluded value, so the solution set contains one number, \(x=−1\), or \(\{−1\}\) written in set notation.

Exercise \(\PageIndex{4}\)

\(\dfrac{2}{3x}=\dfrac{1}{4}-\dfrac{1}{6x}\)

\(x=\dfrac{10}{3}\)

Example \(\PageIndex{5}\): Solving a Rational Equation by Factoring the Denominator

\(\dfrac{1}{x}=\dfrac{1}{10}-\dfrac{3}{4x}\)

First find the common denominator. The three denominators in factored form are \(x,10=2⋅5\), and \(4x=2⋅2⋅x\). The smallest expression that is divisible by each one of the denominators is \(20x\). Only \(x=0\) is an excluded value. Multiply the whole equation by \(20x\).

\[\begin{align*} 20x\left(\dfrac{1}{x}\right)&= \left(\dfrac{1}{10}-\dfrac{3}{4x}\right)20x\\ 20&= 2x-15\\ 35&= 2x\\ \dfrac{35}{2}&= x \end{align*}\]

The solution is \(\dfrac{35}{2}\).

Exercise \(\PageIndex{5}\)

\[-\dfrac{5}{2x}+\dfrac{3}{4x}=-\dfrac{7}{4} \nonumber \]

Example \(\PageIndex{6}\): Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

• \(\dfrac{3}{x-6}=\dfrac{5}{x}\)
• \(\dfrac{x}{x-3}=\dfrac{5}{x-3}-\dfrac{1}{2}\)
• \(\dfrac{x}{x-2}=\dfrac{5}{x-2}-\dfrac{1}{2}\)

The denominators \(x\) and \(x−6\) have nothing in common. Therefore, the LCD is the product \(x(x−6)\). However, for this problem, we can cross-multiply.

\[\begin{align*} \dfrac{3}{x-6}&=\dfrac{5}{x}\\ 3x&=5(x-6)\qquad \text{Distribute.}\\ 3x&=5x-30\\ -2x&=-30\\ x&=15 \end{align*}\]

The solution is \(15\). The excluded values are \(6\) and \(0\).

The LCD is \(2(x−3)\). Multiply both sides of the equation by \(2(x−3)\).

\[\begin{align*} 2(x-3)\left [\dfrac{x}{x-3} \right ]&= \left [\dfrac{5}{x-3}-\dfrac{1}{2} \right ]2(x-3)\\ \dfrac{2(x-3)x}{x-3}&= \dfrac{2(x-3)5}{x-3}-\dfrac{2(x-3)}{2}\\ 2x&= 10-(x-3)\\ 2x&= 13-x\\ 3x&= 13\\ x&= \dfrac{13}{3} \end{align*}\]

The solution is \(\dfrac{13}{3}\). The excluded value is \(3\).

The least common denominator is \(2(x−2)\). Multiply both sides of the equation by \(x(x−2)\).

\[\begin{align*} 2(x-2)\left [\dfrac{x}{x-2} \right ]&= \left [\dfrac{5}{x-2}-\dfrac{1}{2} \right ]2(x-2)\\ 2x&= 10-(x-2)\\ 2x&= 12-x\\ 3x&= 12\\ x&= 4 \end{align*}\]

The solution is \(4\). The excluded value is \(2\).

Exercise \(\PageIndex{6}\)

Solve \(\dfrac{-3}{2x+1}=\dfrac{4}{3x+1}\). State the excluded values.

\(x=-\dfrac{7}{17}\). Excluded values are \(x=−12\) and \(x=−13\).

Example \(\PageIndex{7}\): Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: \(\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{x^2-1}\). State the excluded values.

We must factor the denominator \(x^2−1\). We recognize this as the difference of squares, and factor it as \((x−1)(x+1)\). Thus, the LCD that contains each denominator is \((x−1)(x+1)\). Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.

\[\begin{align*} (x+1)(x-1)\left [\dfrac{2}{x+1}-\dfrac{1}{x-1} \right ]&= \left [\dfrac{2x}{x^2-1} \right ](x+1)(x-1)\\ 2(x-1)-(x+1)&= 2x\\ 2x-2-x-1&= 2x \text{ Distribute the negative sign}\\ -3-x&= 0\\ x&= -3 \end{align*}\]

The solution is \(−3\). The excluded values are \(1\) and \(−1\).

Exercise \(\PageIndex{7}\)

\(\dfrac{2}{x-2}+\dfrac{1}{x+1}=\dfrac{1}{x^2-x-2}\)

\(x=\dfrac{1}{3}\)

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form , written as \[y=mx+b\] where \(m=\text{slope}\) and \(b=\text{y−intercept.}\) Let us begin with the slope.

The slope of a line refers to the ratio of the vertical change in \(y\) over the horizontal change in \(x\) between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure \(\PageIndex{2}\). The lines indicate the following slopes: \(m=−3\), \(m=2\), and \(m=\dfrac{1}{3}\).

THE SLOPE OF A LINE

The slope of a line, \(m\), represents the change in \(y\) over the change in \(x\). Given two points, \((x_1,y_1)\) and \((x_2,y_2)\), the following formula determines the slope of a line containing these points:

\[m=\dfrac{y_2-y_1}{x_2-x_1}\]

Example \(\PageIndex{8}\): Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points \((2,−1)\) and \((−5,3)\).

We substitute the \(y\)-values and the \(x\)-values into the formula.

\[\begin{align*} m&= \dfrac{3-(-1)}{-5-2}\\ &= \dfrac{4}{-7}\\ &= -\dfrac{4}{7} \end{align*}\]

The slope is \(-\dfrac{4}{7}\)

It does not matter which point is called \((x_1,y_1)\) or \((x_2,y_2)\). As long as we are consistent with the order of the \(y\) terms and the order of the \(x\) terms in the numerator and denominator, the calculation will yield the same result.

Exercise \(\PageIndex{8}\)

Find the slope of the line that passes through the points \((−2,6)\) and \((1,4)\).

\(m=-\dfrac{2}{3}\)

Example \(\PageIndex{9}\): Identifying the Slope and y-intercept of a Line Given an Equation

Identify the slope and \(y\)-intercept, given the equation \(y=-\dfrac{3}{4}x-4\).

As the line is in \(y=mx+b\) form, the given line has a slope of \(m=-\dfrac{3}{4}\). The \(y\)-intercept is \(b=−4\).

The \(y\)-intercept is the point at which the line crosses the \(y\)-axis. On the \(y\)-axis, \(x=0\). We can always identify the \(y\)-intercept when the line is in slope-intercept form, as it will always equal \(b\). Or, just substitute \(x=0\) and solve for \(y\).

The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.

\[y−y_1=m(x−x_1)\]

This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

THE POINT-SLOPE FORMULA

Given one point and the slope, the point-slope formula will lead to the equation of a line:

Example \(\PageIndex{10}\): Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope \(m=−3\) and passing through the point \((4,8)\). Write the final equation in slope-intercept form.

Using the point-slope formula, substitute \(−3\) for m and the point \((4,8)\) for \((x_1,y_1)\).

\[\begin{align*} y-y_1&= m(x-x_1)\\ y-8&= -3(x-4)\\ y-8&= -3x+12\\ y&= -3x+20 \end{align*}\]

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Exercise \(\PageIndex{10}\)

Given \(m=4\), find the equation of the line in slope-intercept form passing through the point \((2,5)\).

\(y=4x−3\)

Example \(\PageIndex{11}\): Finding the Equation of a Line Passing Through Two Given Points

Find the equation of the line passing through the points \((3,4)\) and \((0,−3)\). Write the final equation in slope-intercept form.

First, we calculate the slope using the slope formula and two points.

\[\begin{align*} m&= \dfrac{-3-4}{0-3}\\ m&= \dfrac{-7}{-3}\\ m&= \dfrac{7}{3}\\ \end{align*}\]

Next, we use the point-slope formula with the slope of \(\dfrac{7}{3}\), and either point. Let’s pick the point \((3,4)\) for \((x_1,y_1)\).

\[\begin{align*} y-4&= \dfrac{7}{3}(x-3)\\ y-4&= \dfrac{7}{3}x-7\\ y&= \dfrac{7}{3}x-3\\ \end{align*}\]

In slope-intercept form, the equation is written as \(y=\dfrac{7}{3}x-3\)

To prove that either point can be used, let us use the second point \((0,−3)\) and see if we get the same equation.

\[\begin{align*} y-(-3)&= \dfrac{7}{3}(x-0)\\ y+3&= \dfrac{7}{3}x\\ y&= \dfrac{7}{3}x-3\\ \end{align*}\]

We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.

Standard Form of a Line

Another way that we can represent the equation of a line is in standard form . Standard form is given as

\[Ax+By=C\]

where \(A\), \(B\), and \(C\) are integers. The \(x\)- and \(y\)-terms are on one side of the equal sign and the constant term is on the other side.

Example \(\PageIndex{12}\): Finding the Equation of a Line and Writing It in Standard Form

Find the equation of the line with \(m=−6\) and passing through the point \(\left(\dfrac{1}{4},−2\right)\). Write the equation in standard form.

We begin using the point-slope formula.

\[\begin{align*} y-(-2)&= -6\left(x-\dfrac{1}{4}\right)\\ y+2&= -6x+\dfrac{3}{2}\\ \end{align*}\]

From here, we multiply through by \(2\), as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.

\[\begin{align*} 2(y+2)&= \left(-6x+\dfrac{3}{2}\right)2\\ 2y+4&= -12x+3\\ 12x+2y&= -1 \end{align*}\]

This equation is now written in standard form.

Exercise \(\PageIndex{12}\)

Find the equation of the line in standard form with slope \(m=−\dfrac{1}{3}\) and passing through the point \((1,13)\).

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

where \(c\) is a constant. The slope of a vertical line is undefined, and regardless of the \(y\)-value of any point on the line, the \(x\)-coordinate of the point will be \(c\).

Suppose that we want to find the equation of a line containing the following points: \((−3,−5)\),\((−3,1)\),\((−3,3)\), and \((−3,5)\). First, we will find the slope.

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the \(x\)-coordinates are the same and we find a vertical line through \(x=−3\). See Figure \(\PageIndex{3}\).

The equation of a horizontal line is given as

where \(c\) is a constant. The slope of a horizontal line is zero, and for any \(x\)-value of a point on the line, the \(y\)-coordinate will be \(c\).

Suppose we want to find the equation of a line that contains the following set of points: \((−2,−2)\),\((0,−2)\),\((3,−2)\), and \((5,−2)\). We can use the point-slope formula. First, we find the slope using any two points on the line.

\[\begin{align*} m&= \dfrac{-2-(-2)}{0-(-2)}\\ &= \dfrac{0}{2}\\ &= 0 \end{align*}\]

Use any point for \((x_1,y_1)\) in the formula, or use the y-intercept.

\[\begin{align*} y-(-2)&= 0(x-3)\\ y+2&= 0\\ y&= -2 \end{align*}\]

The graph is a horizontal line through \(y=−2\). Notice that all of the y-coordinates are the same. See Figure \(\PageIndex{3}\).

Example \(\PageIndex{13}\): Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points: \((1,−3)\) and \((1,4)\).

The \(x\)-coordinate of both points is \(1\). Therefore, we have a vertical line, \(x=1\).

Exercise \(\PageIndex{13}\)

Find the equation of the line passing through \((−5,2)\) and \((2,2)\).

Horizontal line: \(y=2\)

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, Figure \(\PageIndex{4}\) shows the graphs of various lines with the same slope, \(m=2\).

All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts.

Lines that are perpendicular intersect to form a \(90^{\circ}\) -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is \(−1:m_1⋅m_2=−1\). For example, Figure \(\PageIndex{5}\) shows the graph of two perpendicular lines. One line has a slope of \(3\); the other line has a slope of \(-\dfrac{1}{3}\).

\[\begin{align*} m_1\cdot m_2&= -1\\ 3\cdot \left (-\dfrac{1}{3} \right )&= -1\\ \end{align*}\]

Example \(\PageIndex{14}\): Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: \(3y=−4x+3\) and \(3x−4y=8\).

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

\[\begin{align*} 3y&= -4x+3\\ y&= -\dfrac{4}{3}x+1\\ \end{align*}\]

Second equation:

\[\begin{align*} 3x-4y&= 8\\ -4y&= -3x+8\\ y&= \dfrac{3}{4}x-2 \end{align*}\]

See the graph of both lines in Figure \(\PageIndex{6}\).

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

\[\begin{align*} m_1&=-\dfrac{4}{3}\\ m_2&=\dfrac{3}{4}\\ m_1\cdot m_2&=\left(-\dfrac{4}{3}\right)\left(\dfrac{3}{4}\right)\\ &=-1 \end{align*}\]

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Exercise \(\PageIndex{14}\)

Graph the two lines and determine whether they are parallel, perpendicular, or neither: \(2y−x=10\) and \(2y=x+4\).

Parallel lines: equations are written in slope-intercept form.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

• Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
• Use the slope and the given point with the point-slope formula.
• Simplify the line to slope-intercept form and compare the equation to the given line.

Example \(\PageIndex{15}\): Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a \(5x+3y=1\) and passing through the point \((3,5)\).

First, we will write the equation in slope-intercept form to find the slope.

\[\begin{align*} 5x+3y&= 1\\ 3y&= -5x+1\\ y&= -\dfrac{5}{3}x+\dfrac{1}{3} \end{align*}\]

The slope is \(m=−\dfrac{5}{3}\). The y-intercept is \(13\), but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the \(y\)-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

\[\begin{align*} y-5&= -\dfrac{5}{3}(x-3)\\ y-5&= -\dfrac{5}{3}x+5\\ y&= -\dfrac{5}{3}x+10 \end{align*}\]

The equation of the line is \(y=−\dfrac{5}{3}x+10\). See Figure \(\PageIndex{8}\).

Exercise \(\PageIndex{15}\)

Find the equation of the line parallel to \(5x=7+y\) and passing through the point \((−1,−2)\).

Example \(\PageIndex{16}\): Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to \(5x−3y+4=0\space(−4,1)\).

The first step is to write the equation in slope-intercept form.

\[\begin{align*} 5x-3y+4&= 0\\ -3y&= -5x-4\\ y&= \dfrac{5}{3}x+\dfrac{4}{3} \end{align*}\]

We see that the slope is \(m=\dfrac{5}{3}\). This means that the slope of the line perpendicular to the given line is the negative reciprocal, or \(-\dfrac{3}{5}\). Next, we use the point-slope formula with this new slope and the given point.

\[\begin{align*} y-1&= -\dfrac{3}{5}(x-(-4))\\ y-1&= -\dfrac{3}{5}x-\dfrac{12}{5}\\ y&= -\dfrac{3}{5}x-\dfrac{12}{5}+\dfrac{5}{5}\\ y&= -\dfrac{3}{5}x-\dfrac{7}{5} \end{align*}\]

Access these online resources for additional instruction and practice with linear equations.

• Solving rational equations
• Equation of a line given two points
• Finding the equation of a line perpendicular to another line through a given point
• Finding the equation of a line parallel to another line through a given point

Key Concepts

• We can solve linear equations in one variable in the form \(ax +b=0\) using standard algebraic properties. See Example and Example .
• A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See Example and Example .
• All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See Example and Example .
• Given two points, we can find the slope of a line using the slope formula. See Example .
• We can identify the slope and \(y\)-intercept of an equation in slope-intercept form. See Example .
• We can find the equation of a line given the slope and a point. See Example .
• We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See Example .
• The standard form of a line has no fractions. See Example .
• Horizontal lines have a slope of zero and are defined as \(y=c\), where \(c\) is a constant.
• Vertical lines have an undefined slope (zero in the denominator), and are defined as \(x=c\), where \(c\) is a constant. See Example .
• Parallel lines have the same slope and different \(y\)-intercepts. See Example .
• Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See Example .

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Word Problems Linear Equations

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\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

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SAT Mathematics : Solving Linear Equations

Study concepts, example questions & explanations for sat mathematics, all sat mathematics resources, example questions, example question #1 : solving linear equations.

Example Question #2 : Solving Linear Equations

Example Question #3 : Solving Linear Equations

Example Question #4 : Solving Linear Equations

You can then just flip each fraction (in doing so, you're doing the same thing to both sides, namely taking the reciprocal), to get:

Example Question #5 : Solving Linear Equations

Whenever the SAT asks you to solve for the value of a variable, you have two options: you can either use your answers as assets and backsolve, or you can solve for the variable directly. In this case, both are equally good options, and you should use whichever method is most comfortable. Generally, however, solving directly is usually the best option on the SAT non-calculator section.

Example Question #6 : Solving Linear Equations

Example Question #21 : Linear Algebra

To solve this problem algebraically, distribute the multiplication across each set of parentheses, remembering to multiply each term within the parentheses by its coefficient. That gives you:

Example Question #10 : Solving Linear Equations

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