## WORD PROBLEMS INVOLVING LINEAR EQUATIONS IN TWO VARIABLES

Problem 1 :

Raman’s age is three times he sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman

Let x be Raman's age and y be the sum of ages of his two sons.

x = 3y -----(1)

After 5 years :

Raman's age = x + 5

The sum of the ages of his two sons = y + 5 + 5

= y + 10

x + 5 = 2(y + 10)

x + 5 = 2y + 20

x - 2y = 20 - 5

x - 2y = 15 --------(2)

By applying the value of x in (2), we get

3y - 2y = 15

y = 15

x = 3(15)

x = 45

So, Raman is 45 years old.

Problem 2 :

The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number

The required number will be in the form X 0 Y

Middle digit = 0

x + y = 13 -------(1)

Y 0 X = X0Y + 495

100y + x = 100x + 1y + 495

x - 100x + 100y - y = 495

-99x + 99y = 495

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

x = y - 5

By applying the value of x in (1), we get

y - 5 + y = 13

2y = 13+5

2y = 18

y = 9

When y = 9,

x = 9 - 5

x = 4

So, the required number is 409.

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## How to Solve Two-variable Linear Equations Word Problems

Two-variable Linear Equations Word Problems refer to a specific category of mathematical problems that present a real-life scenario requiring the formulation and subsequent solution of two linear equations with two unknown variables. These problems demand a blend of comprehension skills and algebraic manipulation to arrive at a solution.

## Step-by-step Guide to Solve Two-variable Linear Equations Word Problems

Here is a step-by-step guide to solving two-variable linear equations word problems:

## Step 1: Initial Gathering of Thoughts:

Before anything else, immerse yourself in the story the problem tells. Without rushing to solve, familiarize yourself with the narrative.

## Step 2: Key Element Identification:

Discover and underline essential entities (like quantities or amounts) and the relationships between them. These details will guide the formation of your equations.

## Step 3: Symbolization of Unknowns:

Give a face (or a symbol) to the unknowns. Let’s take, for instance, \(x\) the first unknown and \(y\) for the second.

## Step 4: Sculpting the Equations:

Use the relationships you’ve identified to craft the equations. Remember, the art lies in ensuring that the equations genuinely capture the essence of the story.

## Step 5: Assembly of the System:

Place the two equations side-by-side, creating a picturesque system. This will help you visualize the interconnections.

## Step 6: Selection of the Tackling Technique:

There are multiple ways to unmask the unknowns:

- Substitution Method: Isolate one of the variables and replace it in the other equation.
- Elimination Method: Adjust the coefficients such that adding or subtracting the equations eliminates one of the variables.
- Graphical Method: If you’re visual, sketch both equations on a graph. The point of intersection is your solution.

## Step 7: Decoding the Equations:

Solve the jigsaw! Dive deep into the equations to discover the values of \(x\) and \(y\). Remember, accuracy is paramount.

## Step 8: Intersecting Solutions with the Story:

Revisit the original narrative. Integrate your solutions (the values of \(x\) and \(y\)) back into the story to ensure everything aligns harmoniously.

## Step 9: Reality Check:

Dive back in! Does your solution make sense in the context of the word problem? If your answers seem absurd or out of place, retrace your steps.

## Step 10: Elation & Expression:

Once you’re confident, articulate your solution clearly, with emphasis on each step’s logic. Your journey through the problem is as essential as the solution itself.

Remember, in the cosmos of mathematics, every word problem is a tale waiting for its plot to be unraveled. Your task is not just to solve but to be the storyteller, unveiling each twist with precision and care. Happy solving!

by: Effortless Math Team about 3 months ago (category: Articles )

## Effortless Math Team

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## Mastering Grade 6 Math Word Problems The Ultimate Guide to Tackling 6th Grade Math Word Problems

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## Word Problems Linear Equations

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

## \(\textbf{1)}\) 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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## Writing Systems of Linear Equations from Word Problems

As we learn about math, we often ask ourselves, "How is any of this knowledge useful in life? When will I ever need to use algebra?" Here's the thing: When you know how to "translate" word problems into algebraic equations, you'll immediately see how useful math really is. With a few simple steps, you can turn real questions about the world around you into equations, allowing you to calculate things that you never thought possible. It's one of the most interesting things about math -- so let's get started!

## How to tell when a word problem can become a linear equation

First, we need to keep our eyes open for a number of clues. These clues tell us that we can turn our word problem into a linear equation:

- There are different quantities of things, such as a specific number of people, objects, hours, and so on.
- Each quantity has a clear value. Instead of saying, "There are a few boxes," the word problem needs to tell us how many boxes there are. Are there five? Six? Seven?
- We need to know at least some of these values to build our linear equation. The unknowns can become variables, like "x" or "y."

## How to turn word problems into linear equations

If we follow a few simple steps, we can turn certain word problems into linear equations:

- Take a second to think about the "problem." What are we trying to find out? What is the "variable" in this word problem? Define all of the words carefully. If we can't define the words properly, our equation won't be accurate.
- Turn the word problem into an equation. Plug in all of your known values and use letters like "x" and "y" to represent the unknown variables. Make sure you write out what each variable represents below the equation so you don't forget.
- Solve the equation. Using our math skills, we can now solve the problem and find the values of our variables. We can use a wide range of strategies to solve the equation, including substitution, elimination, or graphing.

## An example of a word problem translated into a linear equation

Now let's see how this all works with an example. Here's our word problem:

We decided to go to a music concert with all our friends, including 12 children and 3 adults. We paid for everyone's tickets for a total of $162. Another group of friends paid $122 for 8 children and 3 adults. How much does a child's ticket cost, and how much does an adult's ticket cost?

1. Understand the problem

We know two values: 12 children and 3 adults cost $162, while 8 children and 3 adults cost $122.

What we don't know is how much an adult's ticket costs, and how much a child's ticket costs.

Let's create variables for those unknowns:

x = the cost of one child's ticket

y = the cost of one adult's ticket

2. Translate the problem into an equation

We know that 12 children and 3 adults cost $162. Let's plug in our variables and create an equation based on this:

12x + 3y = 162

We can do the same for the other group of concert-goers:

8x + 3y = 122

3. Solve the equation

On one weekend they sold a total of 12 adult tickets and 3 child tickets for a total of 162 dollars, and the next weekend they sold 8 adult tickets and 3 child tickets for 122 dollars, find the price for a child's and adult's ticket.

When we have two very similar equations like this, we can simply subtract them from each other to get the values we need:

Now that we know the value of x, we can use it to find y.

12(10) + 3y = 162

120 + 3y = 162

Now we know that a child's ticket costs $10, while an adult's ticket costs $14.

We can now check our work by plugging our solutions back into our original equations:

12(10) + 3(14) = 162

8(10) + 3(14) = 122

You can use this strategy to solve all kinds of everyday math problems you encounter in life!

## Topics related to the Writing Systems of Linear Equations from Word Problems

Consistent and Dependent Systems

Word Problems

Cramer's Rule

## Flashcards covering the Writing Systems of Linear Equations from Word Problems

Algebra 1 Flashcards

College Algebra Flashcards

## Practice tests covering the Writing Systems of Linear Equations from Word Problems

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

## Get more help with linear equation word problems

Does your child need a little extra help? Are they craving new challenges? Contact Varsity Tutors today, and we will find them a professional math tutor whose skills match your student's unique needs. Whether they need more help identifying elements from a word problem to plug into a potential linear equation or they need more challenging problems, a tutor can provide exactly the type of assistance your student needs.

## Download our free learning tools apps and test prep books

## 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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## 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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## Word Problems on Linear equation in one variable

In this chapter we will solve some word problem questions related to linear equation in one variable.

To solve the questions, you have to follow below steps ;

(a) Frame the equation after reading the question. (b) Solve the equation and find the unknown variable. Hence, apart from math skills, you should also posses reading comprehension skills so that you can frame the equation accurately.

To have basic understanding of solving linear equation, click the below links.

How to solve linear equation ? Solving linear equation using cross multiplication

## Word Problems on Linear equation with one variable.

Example 01 For cleaning the garden, Sam was paid 20$. After the payment, Sam has 60$ in his wallet. Find the money in the wallet before the payment?

Solution Let the money before payment was x $ . We know that Sam received $20 for his services and after the payment he has 60$ . We can say that; Initial money + 20 = 60 x + 20 = 60 x = 60 – 20 x = 40 Hence, Sam had 40$ before the payment.

Example 02 Peter is 5 years younger than Ashley. Four years later, Ashley will be twice as old as Peter. Find their present ages?

Solution Let Ashley’s age be x . Then Peter age will be x – 5.

Age of both the person after 4 years. Ashley = x + 4 Peter = x – 5 + 4 It is said that after 4 years, Ashley age is twice as old as Peter. So we multiply Peter age by 2 to make it equal to Ashley’s age . 2 . (x – 5 + 4) = x + 4 2 (x – 1) = x + 4

2x – 2 = x + 4 2x – x = 4 + 2 x = 6 Hence, Ashley Present age is 6 years . Peter Present age is = 6 – 5 + 4 = 5 years

Example 03 If we multiply weight of the box by 8/3, we will get 24 Kg. What is the original weight of the box.

Solution Let weight of the box be x kg . According to question, if we multiply the box by 8/3 we will get 24 kg.

Solving the linear equation, we get;

Hence, the original weight of box is 9 Kg.

Example 04 When number is divided by 8, the result is 5. Find the original number.

Solution Let the original number be x.

According to question, when we divide number x by 8 we get number 5 .

Hence, 40 is the original number.

Example 05 The sum of two consecutive multiple of 6 is 54. Find both the numbers.

Solution Let the first multiple of 6 be x . The other consecutive multiple of 6 will be x + 6.

According to question, both the numbers will add to number 54.

The linear equation is written as;

x + (x + 6) = 54

Solving the linear equation. x + x + 6 = 54 2x = 54 – 6 2x = 48 x = 48 / 2 x = 24.

The two original numbers are; x = 24 x + 6 = 24 + 6 = 30

Hence, number 24 & 30 are the right solution .

Example 06 The sum of two numbers is 15 and their difference is 7. Find the original numbers.

Solution Let the two numbers be x & y. It’s given that sum of two numbers is 15 . x + y = 15 y = 15 – x So, the two numbers are x and (15 – x).

According to question, the difference of these two numbers is 7. x – (15 – x ) = 7 x – 15 + x = 7 2x = 7 + 15 2x = 22 x = 11

So the two numbers are; x = 11 15 – x = 15 – 11 = 4 Hence, 11 & 4 are the solution .

Example 07 The angles of triangle are in ratio 1: 2: 3. Find the measure of all the angles. Solution Let the angles of triangle are x, 2x and 3x .

We know that sum of all angle of triangle measures 360 degree .

x + 2x + 3x = 360 6x = 360 x = 360 / 6 x = 60 degree. So the measure of all angles are; x = 60 degree 2x= 2 (60) = 120 degree 3x = 3 (60) = 180 degree

Hence, 60, 120 and 180 degree is the solution.

Example 08 The cost of three laptops and two printer is 1100$. If the Laptop costs 200$ more than printer, then find the price of both the items.

Solution Let the laptop price be x and printer price be y . According to question, laptop cost 200$ more than printer. So; Price of Laptop = 200 + price of printer x = 200 + y y = x – 200 Now the laptop price is x and printer price is x – 200 .

According to question, 3 Laptop and 2 printer cost 1100$ 3x + 2 (x – 200) = 1100 3x + 2x – 400 = 1100 5x = 1100 + 400 5x = 1500 x = 1500 / 5 x = 300 So; Price of laptop ( x ) = 300$ Price of printer = 300 – 200 = 100$

## Related posts:

- Problems on Linear equation in one variable
- Cross multiplication method for linear equation
- What is Linear Equation?
- Solving linear equations
- Linear Inequality
- Compound Interest Solved Problems
- Subtraction Word Problems || Grade 1 Math Quiz
- Graphing Linear Function
- Solving systems of equation by graphing
- Solving linear inequality

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p=1/5 and q=1/11. Hence x−y =5 x − y = 5 and x +y = 11 x + y = 11, Solving these, x=8 km/hr and y=3km/hr. Question 8. A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in 6 hrs and 30 minutes. Find the speed of the boat in still water and also speed of the stream. Hint:

Solution: Adding 9 to a number is written as x + 9, while subtracting 7 from three times the number is written as 3x − 7. We therefore get the equation: x + 9 = 3x − 7. We solve for x by adding 7 on both sides of the equation: x + 16 = 3x. Then we subtract x: 16 = 2x. After dividing by 2, we obtain the answer x = 8.

WORD PROBLEMS INVOLVING LINEAR EQUATIONS IN TWO VARIABLES. Raman's age is three times he sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman. Let x be Raman's age and y be the sum of ages of his two sons. So, Raman is 45 years old.

Word problems that require us to write systems of linear equations have two unknown quantities and two different ways to relate them. This means we need to write two linear equations, and each contains the two unknowns as variables. The Understanding linear relationships lesson details how to translate verbal descriptions into equations.

Two-variable Linear Equations Word Problems refer to a specific category of mathematical problems that present a real-life scenario requiring the formulation and subsequent solution of two linear equations with two unknown variables. These problems demand a blend of comprehension skills and algebraic manipulation to arrive at a solution.

In Summary. 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.

Word problems involving pair of linear equations (advanced) Jake's and his son's present ages are j years and s years respectively. Jake was 5 times as old as his son 2 years ago. 2 years later, Jake's age will be 8 years more than three times his son's age.

Two-step equations word problems. Mindy and Troy combined ate 9 pieces of the wedding cake. Mindy ate 3 pieces of cake and Troy had 1 4 of the total cake. Write an equation to determine how many pieces of cake ( c) there were in total.

Turn the word problem into an equation. Plug in all of your known values and use letters like "x" and "y" to represent the unknown variables. Make sure you write out what each variable represents below the equation so you don't forget. Solve the equation.

Unit 1 Real numbers. Unit 2 Polynomials. Unit 3 Pair of linear equations in two variables. Unit 4 Quadratic equations. Unit 5 Arithmetic progressions. Unit 6 Triangles. Unit 7 Coordinate geometry. Unit 8 Introduction to trigonometry. Unit 9 Some applications of trigonometry.

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 ...

http://www.greenemath.com/In this video we demonstrate how to set up and solve a word problem that involves a linear equation in two variables.

Word Problems: Linear Equations in Two Variables. 1. The perimeter of a rectangular garden is 20 m. If the length is 4 m more than the breadth, find the length and breadth of the garden. Solution: Let the length of the garden be x m. Therefore, breadth of garden = (x - 4) m. Hence, length = 7 m and breadth = 7 - 4 = 3 m.

9=9 (9+9=18) divide 4 by 2, you get 2. (4 is the difference between the two values) now add 2 to one side, and subtract 2 from the other. 9+2=9-2. 11=7 (11+7=18) Some of you guys will overthink something as simple as this, read the problem carefully and evaluate what you have learned one by one to successfully get the answer.

Problems Involving Relationships between Two Variables. If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.

Don't Memorise brings learning to life through its captivating FREE educational videos. To Know More, visit https://InfinityLearn.comNew videos every week. T...

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. Let the number be x. 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.

Writing two-variable inequalities word problem. ... Given a two-variable linear inequality that models a context about watering plants, Sal finds how many flowers can be watered. ... Well, flipping 7/10 to 10/7 to multiply the equations is called multiplying by the reciprocal or inverse (of the fraction). 10/7 is the reciprocal or inverse of 7/ ...

Section 7.1 : Linear Systems with Two Variables. For problems 1 - 3 use the Method of Substitution to find the solution to the given system or to determine if the system is inconsistent or dependent. x−7y = −11 5x+2y = −18 x − 7 y = − 11 5 x + 2 y = − 18 Solution.

In this comprehensive lesson, Ratan Sir unravels the essence of linear equations in two variables and how they relate to real-life scenarios. You'll gain a s... CBSE Exam, class 10

Lesson 1- Solving Problems involving Linear Equation in Two Variables After going through this module, you are expected to: 1. translate word problems into linear equations; 2. solve problems involving linear equations in two variables; and 3. apply the concepts of solving linear equations in dealing with real-life problems.

Course: Algebra 1 > Unit 7. Lesson 3: Modeling with linear inequalities. Writing two-variable inequalities word problem. Solving two-variable inequalities word problem. Interpreting two-variable inequalities word problem. Two-variable inequalities word problems. Modeling with systems of inequalities. Writing systems of inequalities word problem.

Understand the concept of Word Problems - Linear Equation in Two Variables with CBSE Class 10 course curated by Kartik Gautam on Unacademy. The Mathematics course is delivered in Hinglish. ... In this class, Kartik Gautam will discuss Pair of Linear Equation in Two Variables. It will be helpful for the aspirants preparing for Class 10.

In this chapter we will solve some word problem questions related to linear equation in one variable. ... Word Problems on Linear equation with one variable. Example 01 For cleaning the garden, Sam was paid 20$. After the payment, Sam has 60$ in his wallet. ... the difference of these two numbers is 7. x - (15 - x ) = 7 x - 15 + x = 7 2x ...