solving linear equations in one variable practice problems

2.2 Linear Equations in One Variable

Learning objectives.

In this section, you will:

• 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 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 a x + b = 0 a x + 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.

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 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 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following:

The solution set consists of one number: { − 4 } . { − 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 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following:

Indeed, −15 ≠ −20. −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. a ≠ 0.

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 = _________, 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 ) = a b + a c . a ( b + c ) = a b + a c .
• 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.

Solving an Equation in One Variable

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

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

The solution is 6.

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

Solving an Equation Algebraically When the Variable Appears on Both Sides

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

Apply standard algebraic properties.

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 .

Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 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 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

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.

Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 .

We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x 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 6 x . 6 x .

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 ) . ( 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 , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 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 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( 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 , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following:

So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( 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 x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . 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.

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 ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c .

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

Rational Equations

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

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.

Solving a Rational Equation without Factoring

Solve the following rational equation:

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

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

Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x .

Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x .

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

The solution is 35 2 . 35 2 .

Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 .

Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

• ⓐ 3 x − 6 = 5 x 3 x − 6 = 5 x
• ⓑ x x − 3 = 5 x − 3 − 1 2 x x − 3 = 5 x − 3 − 1 2
• ⓒ x x − 2 = 5 x − 2 − 1 2 x x − 2 = 5 x − 2 − 1 2

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

The solution is 15. The excluded values are 6 6 and 0. 0.

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

The solution is 13 3 . 13 3 . The excluded value is 3. 3.

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

The solution is 4. The excluded value is 2. 2.

Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values.

Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values.

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

The solution is −3. −3. The excluded values are 1 1 and −1. −1.

Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 .

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope.

The Slope of a Line

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 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 .

The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points:

Finding the Slope of a Line Given Two Points

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

We substitute the y- values and the x- values into the formula.

The slope is − 4 7 . − 4 7 .

It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( 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.

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

Identifying the Slope and y- intercept of a Line Given an Equation

Identify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4.

As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4.

The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. 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 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.

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.

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

Finding the Equation of a Line Given the Slope and One Point

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

Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

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

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

Finding the Equation of a Line Passing Through Two Given Points

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

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

Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3.

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

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

where A , A , B , B , and C 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.

Finding the Equation of a Line and Writing It in Standard Form

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

We begin using the point-slope formula.

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.

This equation is now written in standard form.

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

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 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −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. x = −3. See Figure 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 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line.

Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept.

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

Finding the Equation of a Line Passing Through the Given Points

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

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

Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 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 4 shows the graphs of various lines with the same slope, m = 2. 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° 90° -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. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 .

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: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8.

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

First equation:

Second equation:

See the graph of both lines in Figure 6

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

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

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

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.

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

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

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

The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , 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.

The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 .

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

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) .

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

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

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

2.2 Section Exercises

What does it mean when we say that two lines are parallel?

What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed?

What does it mean when we say that a linear equation is inconsistent?

When solving the following equation:

2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1

explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set.

For the following exercises, solve the equation for x . x .

7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9

4 x − 3 = 5 4 x − 3 = 5

3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 )

12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5

1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3

x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12

2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6

3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3

2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4

x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2

For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set.

3 x − 1 3 = 1 6 3 x − 1 3 = 1 6

2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4

3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 )

3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1

5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3

1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x

For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.

( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3

( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5

x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 )

y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 )

( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 )

( 1 , 3 )  and   ( 5 , 5 ) ( 1 , 3 )  and   ( 5 , 5 )

parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 )

perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) .

For the following exercises, find the equation of the line using the given information.

( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 )

( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 )

The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) .

The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) .

The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) .

( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 )

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4

3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6

y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2

x = 4 y = −3 x = 4 y = −3

For the following exercises, find the slope of the line that passes through the given points.

( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 )

( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 )

( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 )

( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 )

( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 )

For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 ) ( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 )

( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 ) ( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 )

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.

0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100

4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528

200 − 30 y x = 70 200 − 30 y x = 70

Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m .

Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form.

Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25.

Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.

( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 )

Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

The slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.

If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and   x = 75. p = $ 453 and   x = 75.

For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled.

What is your cost if you travel 50 mi?

If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling?

Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

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• Book title: College Algebra 2e
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• Section URL: https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable

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• Linear(Simple) Equations

Linear(Simple) Equations: Very Difficult Problems with Solutions

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Solving linear equations with one variable.

Linear equations in one variable are equations where the variable has an exponent of 1, which is typically not shown (it is understood). An example would be something like \(12x = x – 5\). To solve linear equations, there is one main goal: isolate the variable . In this lesson, we will look at how this is done through several examples.

Table of Contents

• Examples of solving one-step equations
• Examples of solving two-step equations
• Examples of equations where you must simplify first
• Infinitely many or no solutions

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Examples of solving one-step linear equations

After all your hard work solving the equation, you know that you want a final answer like \(x=5\) or \(y=1\). In both of these cases the variable is isolated , or by itself.

So we need to figure out how to isolate the variable. How we do this depends on the equation itself! If it was multiplied by something, we will divide. If something was added to it, we will subtract. By doing this, we will slowly be getting the variable by itself.

Let’s use an example to see how this works.

Solve the equation: \(4x = 8\)

In this example, the 4 is multiplying the \(x\). Therefore, to isolate \(x\), you must divide that side by 4. When doing this, you must remember one important rule: whatever you do to one side of the equation, you must do to the other side. So we will divide both sides by 4.

\(\begin{align}4x &= 8 \\ \dfrac{4x}{\color{red}{4}} &= \dfrac{8}{\color{red}{4}}\end{align}\)

Simplifying:

\(x = \boxed{2}\)

That’s it, one step and we are done. (That’s why equations like these are often called “one-step” equations)

Anytime you are solving linear equations, you can always check your answer by substituting it back into the equation. If you get a true statement, then the answer is correct. This isn’t 100% necessary for every problem, but it is a good habit so we will do it for our equations.

In this example, our original equation was \(4x = 8\). To check this, verify the following is true:

\(\begin{align}4x &= 8\\ 4(2) &= 8 \\ 8 &= 8\end{align}\)

This is a true statement, so our answer is correct.

For any equation, whatever operation you do to one side must also be done to the other side

Let’s try a couple more examples before moving on to more complex equations.

Solve: \(3x=12\)

Since \(x\) is being multiplied by 3, the plan is to divide by 3 on both sides:

\(\begin{align}3x &=12\\ \dfrac{3x}{\color{red}{3}} &=\dfrac{12}{\color{red}{3}}\\ x&= \boxed{4}\end{align}\)

To check our answer, we will let \(x = 4\) and substitute it back into the equation:

\(\begin{align}3x &= 12\\3(4) &= 12 \\ 12 &= 12\end{align}\)

Just as before, since this is a true statement, we know our answer is correct.

In the next example, instead of the variable being multiplied by a value, a value is being subtracted from the variable. To “undo” this, we will add that value to both sides.

Solve: \(y-9=21\)

This time, 9 is being subtracted from y. So, we will undo that by adding 9 to both sides.

\(\begin{align}y-9&=21\\ y-9 \color{red}{+9}&=21\color{red}{+9}\\y&=30\end{align}\)

Next we will look at what are commonly called “two-step” equations. In these equations, we will need to undo two operations in order to isolate the variable.

Examples of two step equations

In each of the examples above, there was a single step to perform before we had our answer. In these next examples, you will see how to work with equations that have two steps instead. If there is more than one operation, it is important to remember the order of operations, PEMDAS . Since you are undoing the operations to \(x\), you will work from the “outside in”. This is easier to understand when you see it in an example.

Solve: \(2x-7=13\)

Notice the two operations happening to \(x\): it is being multiplied by 2 and then having 7 subtracted. We will need to undo these. But, only the \(x\) is being multiplied by 2, so the first step will be to add 7 to both sides. Then we can divide both sides by 2.

Adding 7 to both sides:

\(\begin{align} 2x-7 &= 13\\ 2x-7 \color{red}{+7} & =13 \color{red}{+7}\\ 2x&=20\end{align}\)

Now divide both sides by 2:

\(\begin{align} 2x &=20 \\ \dfrac{2x}{\color{red}{2}}&=\dfrac{20}{\color{red}{2}}\\ x&= \boxed{10}\end{align}\)

Just like with simpler problems, you can check your answer by substituting your value of \(x\) back into the original equation.

\(\begin{align}2x-7&=13\\ 2(10) – 7 &= 13\\ 13 &= 13\end{align}\)

This is true, so we have the correct answer.

Let’s look at one more two-step example before we jump up in difficulty again. Make sure that you understand each step shown and work through the problem as well.

Solve: \(5w + 2 = 9\)

As above, there are two operations: \(w\) is being multiplied by 5 and then has 2 added to it. We will undo these by first subtracting 2 from both sides and then dividing by 5.

\(\begin{align}5w + 2 &= 9\\ 5w + 2 \color{red}{-2} &= 9 \color{red}{-2}\\ 5w &= 7\\ \dfrac{5w}{\color{red}{5}} &=\dfrac{7}{\color{red}{5}}\\w=\boxed{\dfrac{7}{5}}\end{align}\)

The fraction on the right can’t be simplified, so that is our final answer.

Let \(w = \dfrac{7}{5}\). Then:

\(\begin{align}5w + 2 &= 9\\ 5\left(\dfrac{7}{5}\right) + 2 &= 9\\ 7 + 2 &= 9\\ 9 &= 9 \end{align}\)

So, we have the correct answer once again!

Simplifying before solving

In the following examples, there are more variable terms and possibly some simplification that needs to take place. In each case, the steps will be to first simplify both sides, then use what we have been doing to isolate the variable. We will first take an in depth look at an example to see how this all works.

To understand this section, you should be comfortable with combining like terms .

Solve: \(3x+2=4x-1\)

Since both sides are simplified (there are no parentheses we need to figure out and no like terms to combine), the next step is to get all of the x’s on one side of the equation and all the numbers on the other side. The same rule applies – whatever you do to one side of the equation, you must do to the other side as well!

It is possible to either move the \(3x\) or the \(4x\). Suppose you moved the \(4x\). Since it is positive, you would do this by subtracting it from both sides:

\(\begin{align}3x+2 &=4x-1\\ 3x+2\color{red}{-4x} &=4x-1\color{red}{-4x}\\ -x+2 & =-1\end{align}\)

Now the equation looks like those that were worked before. The next step is to subtract 2 from both sides:

\(\begin{align}-x+2\color{red}{-2} &= -1\color{red}{-2}\\-x=-3\end{align}\)

Finally, since \(-x= -1x\) (this is always true), divide both sides by \(-1\):

\(\begin{align}\dfrac{-x}{\color{red}{-1}} &=\dfrac{-3}{\color{red}{-1}}\\ x&=3\end{align}\)

You should take a moment and verify that the following is a true statement:

\(3(3)+ 2 = 4(3) – 1\)

In the next example, we will need to use the distributive property before solving. It is easy to make a mistake here, so make sure that you distribute the number in front of the parentheses to all the terms inside.

Solve: \(3(x+2)-1=x-3(x+1)\)

First, distribute the 3 and –3, and collect like terms.

\(\begin{align} 3(x+2)-1 &=x-3(x+1)\\ 3x+6-1&=x-3x-3 \\ 3x+5&=-2x-3\end{align}\)

Now we can add 2x to both sides. (Remember you will get the same answer if you instead subtracted 3x from both sides)

\(\begin{align} 3x+5\color{red}{+2x} &=-2x-3\color{red}{+2x}\\ 5x+5& =-3\end{align}\)

From here, we can solve as we did with other two-step equations.

\(\begin{align}5x+5\color{red}{-5} &=-3\color{red}{-5}\\ 5x &=-8\\ \dfrac{5x}{\color{red}{5}}&=\dfrac{-8}{\color{red}{5}}\\ x &= \dfrac{-8}{5} \\ &=\boxed{-\dfrac{8}{5}}\end{align}\)

This was a tough one, so remember to check your answer and make sure no mistake was made. To do that, you will be making sure that the following is a true statement:

\(3\left(-\dfrac{8}{5}+2\right)-1=\left(-\dfrac{8}{5}\right)-3\left(-\dfrac{8}{5}+1\right)\)

(Note: it does work – but you have to be really careful about parentheses!)

Infinitely many solutions and no solutions

There are times when you follow all of these steps and a really strange solution comes up. For example, when solving the equation \(x+2=x+2\) using the steps above, end up with \(0=0\). This is certainly true but what good does it do?

If you get a statement such as this, it means that the equation has infinitely many solutions. Any \(x\) you could think of would satisfy the equation \(x+2=x+2\). The appropriate answer in this case is “infinitely many solutions”.

The other situation comes up when you simplify an equation down into a statement that is never true such as \(3=4\) or \(0=1\). This happens with the equation \(x+5=x-7\) which will lead to \(5= -7\), something that is certainly never true. This means that no \(x\) would satisfy this equation. In other words “no solution”. In summary:

• If you get a statement that is always true like \(5 = 5\) or \(0 = 0\), then there are infinitely many solutions.
• If you get a statement that is always false like \(10 = 11\) or \(1 = 5\), then there are no solutions.

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Solving linear equations is all about isolating the variable. Depending on the equation, this may take as little as one step or many more steps. Always check if you need to simplify one or both sides of the equation first, and always check your answer.

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Linear Equation in One Variable

Linear equation in one variable is the basic equation used to represent and solve for an unknown quantity. It can be easily represented graphically and it is always a straight line. The linear equation is an easy way of representing a math statement. Any variable or symbol can be used to represent unknown quantities but generally, a variable 'x' is used to represent the unknown quantity in the linear equation in one variable. Solving a linear equation includes a set of simple methods. The variables are isolated on one side of the equation and the constants are isolated to another side of the equation, to obtain the final value of the unknown quantity.

What is Linear Equation in One Variable?

Before learning about linear equation in one variable, let us quickly go through the meaning of linear equations . A linear equation is a type of equation in which the degree of each variable in the equation is exactly equal to one. Linear equations in one variable are those equations in which there is only one variable present, and there is only one solution of the equation. When it is drawn on a graph, it appears to be a straight line either horizontally or vertically.

A linear equation in one variable is of the form of ax+b=0, where a and b are any two integers and x is an unknown variable having only one solution. Let us understand this by taking an example- "4 added to a certain number gives 10". Find that number. How can we represent this problem in a simpler way? We can say, x + 4 = 10 , find x. We assigned a variable to that number; this is called an equation. It helps us write large problems like this in a shorter way. This equation has one variable which is x and the highest power of x is one. These kinds of equations are known as linear equations in one variable because the degree of the variable x is one.

Solving Linear Equation in One Variable

The general form of a linear equation in one variable is Ax + B = 0. Here A is the coefficient of x, x is the variable, and B is the constant term. The coefficient and the constant term should be segregated to find the final solution of this linear equation.

Now, let us look at how to solve a one-variable linear equation. An equation is like a weighing balance with equal weights on both sides.

If we add or subtract the same number from both sides of an equation, it still holds. Similarly, if we multiply or divide the same number into both sides of an equation, it still holds. Consider the equation, 3x-2=4. We will perform mathematical operations on the LHS and the RHS so that the balance is not disturbed. Let's add 2 on both sides to reduce the LHS to 3x. This will not disturb the balance. The new LHS is 3x-2+2=3x and the new RHS is 4+2=6. Now let's divide both sides by 3 to reduce the LHS to x. Thus, we have 3x/3=6/3. Hence we have x = 2.

The above steps to solve linear equations in one variable can be summarized in the below-listed points.

• Step-1: Keep the variable term on one side and constants on another side of the equation by adding or subtracting on both sides of the equation.
• Step-2: Simplify the constant terms.
• Step-3: Isolate the variable on one side by multiply or dividing it into both sides of the equation.
• Step-4: Simplify and write the answer.

Linear Equation in One Variable vs Non-Linear Equations

Apart from linear equations in one variable, we have other non-linear equations, which have numerous applications in geometry , trigonometry , and calculus. Linear equations in one variable are single-degree equations and are represented as a line on a coordinate plane. A non-linear equation on the other hand is a curve or non-linear representation on the coordinate axis. A nonlinear equation is of a higher degree. Few examples of non-linear equations are the equation of curves such as a circle , parabola , ellipse , hyperbola .

Some of the samples of linear equations are x = 5, 3x + 7 = 9, 4x + 2y = 11. And some of the examples of non-linear equations are the equation of a circle - x 2 + y 2 = 25, equation of a elipse - x 2 /9 + y 2 /16= 1, equation of a hyperbola - x 2 /16 - y 2 / 25 =1.

Important Notes

The following points help us in clearly summarizing the concepts involved in linear equations in one variable.

• The degree of the variable in linear equations should be exactly equal to one.
• The graph of a linear equation in one variable is a straight line, either horizontal or vertical.
• The solution of a linear equation in one variable is unaffected if any number is added, subtracted, multiplied, or divided on both sides of the equation.

Related Topics

Check the articles given below related to the concept of linear equations in one variable.

• One-Variable Linear Equations and Inequations
• Linear Equations in Two Variables
• Applications of Linear Equations
• Solving Linear Equations
• Graphing Linear Equations
• Linear Equations

Linear Equation in One Variable Examples

Example 1: Twenty years ago, Mikkel's age was one-third of what it is now. What is Mikkel's present age?

We can write the given information by using a linear equation in one variable. Let Mikkel's present age be x years. Twenty years ago, Mikkel's age was (x-20) years. According to the given information, x - 20= x/3 3(x - 20)= x 3x - 60= x 3x - x= 60 2x= 60 x= 60/2 x= 30 Therefore the present age of Mikkel is 30 years.

Example 2: David worked as a stenographer. In June, he was paid $50 per day. However, $10 per day was deducted for the days he remained absent. He received $900 for the number of days he worked. How many days did he work?

Let the number of days he worked be x days. Hence, the number of days he did not work will be = 30 - x. He was paid $50 for each day he worked and $10 was deducted for each day he did not work. At the end of the month, he received $900. According to the given information, we can form a linear equation in one variable as, 50(x) - 10(30 - x) = 900 50x - 300 + 10x = 900 60x = 900 + 300 x = 1200/60 x= 20 Therefore David worked for 20 days.

Example 3: Find the perimeter of the square whose side length x units is given in the form of an equation as 2x/3- 5/6=0

Given, the side-length of the square is 2x/3- 5/6=0. It is in the form of a linear equation in one variable which is x. First, we need to solve this equation to find the value of x. 2x/3- 5/6=0 2x/3- 5/6 + 5/6=0+ 5/6 (adding 5/6 on both the sides) 2x/3= 5/6 x = 5/6 × 3/2 x = 5/4 units Hence, the side length of the square is 5/4 units. Now, to find the perimeter of the square, we need to multiply the side length by 4. The perimeter of a square = 5/4 × 4=5 units. Therefore the perimeter of the square is 5 units.

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Practice Questions on Linear Equation in One Variable

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FAQs on Linear Equation in One Variable

What is a linear equation in one variable with example.

Linear equation in one variable is of the form ax + b = 0. The linear equation in one variable are equations in which the highest degree of every term is one, there is one possible solution of the equation and there is only one variable present in it. An example of a linear equation in one variable is 3y+2=0.

What is the Power of Variable in Linear Equation in One Variable?

The power of the variable in a one-variable linear equation is 1. Referring to an example, 3a + 4 = 11, the power of the variable 'a' is 1.

Can a Linear Equation Have More than One Variable?

Yes, linear equations can have more than one variable. We call such equations linear equations in two variables or linear equations in three variables. The linear equations in two variables are of the form ax + by + c = 0 and a linear equation in three variables is of the form ax + by + cz + d = 0. Here x, y, z are the variables, and a, b and c are coefficients and d is a constant. These linear equation are prominently used in linear programming to find the optimal solutions.

How do you Solve Linear Equations in One Variable?

Steps to solve linear equations in one variable are listed below:

• Keep the variable term on one side and constants to another side of the equation by adding or subtracting on both sides of the equation.
• Simplify the constant terms.
• Isolate the variable on one side by multiply or dividing the same term to both sides of the equation.
• Simplify and write the answer.

How do you Solve Linear Equation in One Variable With Variables on Both Sides?

To solve a linear equation in one variable with variables on both sides, we first bring all the terms with variables on one side and the constants on the other side of the equation. Then, we simplify the equation, isolate the variable and write the final answer of the equation. Let us look at a simple equation to understand this, 4x + 1 = 2x + 7 ⇒ 4x - 2x = 7 - 1 ⇒ 2x = 6 ⇒ x = 6/2 ⇒ x = 3.

What are the Steps in Solving Linear Equations in One Variable?

The most general rule for solving linear equations is that we can add, subtract, multiply or divide the same term into both sides of the equations so that we can find the value of the variable present in it. The steps to solve one-variable linear equations are given below:

• Isolate the variable term on one side by adding or subtracting the same term on both sides of the equation.
• Multiply or divide both sides with the coefficient of the variable so that we can have only the variable on one side of the equation.
• Simplify and get the answer in the form of "x=c".

What is the General Form of a linear equation in One Variable?

The general form of the linear equations in one variable is Ax + B = 0 where x is the variable and A is the coefficient of x and b is the constant term.

• Math Article
• Linear Equation In One Variable

Linear Equation in One Variable

The linear equations in one variable  is an equation which is expressed in the form of ax+b = 0, where a and b are two integers, and x is a variable and has only one solution. For example, 2x+3=8 is a linear equation having a single variable in it. Therefore, this equation has only one solution, which is x = 5/2. Whereas if we speak about linear equation in two variables, it has two solutions.

The concept of linear equation in one variable has been covered in this lesson, including its definition, solutions, examples, word problems and worksheet questions. This is an important topic for Class 6, 7 and 8 students. The concepts covered in this lesson are mentioned below in the table of contents. So, what is one variable equation?

Table of Content:

Linear Equation in One Variable Definition

A linear equation in one variable is an equation which has a maximum of one variable of order 1. It is of the form ax + b = 0 , where x is the variable.

This equation has only one solution. A few examples are:

Standard Form of Linear Equations in One Variable

The standard form of linear equations in one variable is represented as:

• ‘a’ and ‘b’ are real numbers.
• Both ‘a’ and ‘b’ are not equal to zero.

Thus, the formula of linear equation in one variable is ax + b = 0.

Solving Linear Equations in One Variable

For solving an equation having only one variable, the following steps are followed

• Step 1 : Using LCM, clear the fractions if any.
• Step 2: Simplify both sides of the equation.
• Step 3: Isolate the variable.
• Step 4: Verify your answer.

Example of Solution of Linear Equation in One Variable

Let us understand the concept with the help of an example.

For solving equations with variables on both sides, the following steps are followed:

Consider the equation: 5x – 9 = -3x + 19

Step 1: Transpose all the variables on one side of the equation. By transpose, we mean to shift the variables from one side of the equation to the other side of the equation. In the method of transposition, the operation on the operand gets reversed.

In the equation 5x – 9 = -3x + 19, we transpose -3x from the right-hand side to the left-hand side of the equality, the operation gets reversed upon transposition and the equation becomes:

5x – 9 +3x = 19

⇒ 8x -9 = 19

Step 2: Similarly transpose all the constant terms on the other side of the equation as below:

⇒ 8x = 19 + 9

Step 3: Divide the equation with 8 on both sides of the equality.

8x/8 = 28/8

If we substitute x = 28/8 in the equation 5x – 9 = -3x + 19, we will get 9 = 9, thereby satisfying the equality and giving us the required solution.

Related Topics:

• Application of linear equations
• Linear Equations Formula
• Graphing Of Linear Equations
• Linear Equations In Two Variables Class 9
• Important Questions Class 8 Maths Chapter 2 Linear Equations One Variable
• Linear Equations One Variable Worksheet

Linear Equation in One Variable Examples

Example 1 : Solve for x, 2x – 4 = 0

Add 4 both sides

2x – 4 + 4 = 0 + 4

Divide each side by 2, we get

x = 4/2 = 2

So, x = 2 is the answer.

Example 2: Solve 12m – 10 = 6

12m – 10 = 6

Add 10 both sides

12m – 10 + 10 = 6 + 10

Divide each side by 12, we get

12m/12 = 16/12

m = 16/12 = 4/3

Answer: m = 4/3

Linear Equations in One Variable Word Problems

Problem: The length of the legs of an isosceles triangle is 4 meters more than its base. If the Perimeter of the triangle is 44 meters, find the lengths of the sides of the triangle.

Let us assume the base measures ‘x’ meter. Hence, each of the legs measure y = (x + 4) meters.

The Perimeter of a triangle is the sum of the three sides.

The equations are formed and solved as follows:

x + 2(x + 4) = 44

x + 2x + 8 = 44

3x + 8 = 44

3x = 44 – 8 = 36

The length of the base is solved as 12 meters. Hence, each of the two legs measure 16 meters.

Linear Equations in One Variable Word Questions (Worksheet)

A few practice questions are given below.

• Question 1: Solve ( 10x – 7) = 21
• Question 2: Find the multiples, if the sum of two consecutive multiples of 6 is 68.
• Question 3: Verify that if x = -3, is a solution of the linear equation 10x + 7 = 13 – 5x.

Frequently Asked Questions

How many solutions does a linear equation in one variable have.

Every linear equation in one variable has a one and unique solution. If the equation has two or more variables then it becomes a linear equation in two variables or linear equations in three variables and so on and the number of solutions varies as per the count of variables an equation contains.

What is the formula of linear equation in one variable?

The formula or the standard form of an equation having only 1 variable is given as ax + b = 0. In this, there is only 1 variable, i.e. x.

How to easily solve any equation having one variable?

First, put the variable on the left-hand side and the numerical values on the right-hand side. Change the operators while changing sides of the terms and then solve for the variable.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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Worksheet on Linear Equation in One Variable

Practice the questions given in the worksheet on linear equation in one variable.

I. Solve the following equations and check the result:

1. 2x – 6 = 0

2. 7x – 2 = 8 - x

3. 8 – 2x = 5 – 4x

4. 4 + 3x = 2 – 2x

5. 2(x – 1) + 2(3x -1) = 0

6. 4(x-1) –(2x – 5) = 4

7. 2x – \(\frac{2}{3}\) = \(\frac{3}{4}\) – x

8. \(\frac{x}{2}\) – 2= 4 + \(\frac{1}{4}\)

9. \(\frac{x}{9}\) + \(\frac{x}{3}\) = \(\frac{1}{9}\)

10. 3x + 2(x+2) = 10 – (2x – 5)

11. 10(y – 4) – 2(y – 9) – 5(y + 4) = 0

12. \(\frac{2m+5}{3}\) = 3m - 5

13. 6(3x + 2) – 5(6x – 1) = 2(x – 8) – 5(7x – 6) + 8x

14. t – (2t +5) – (1- 2t) = (3 + 4t) – 2(t - 4)

15. \(\frac{2x}{3}\) = \(\frac{3x}{4}\) + \(\frac{7}{12}\)

16. \(\frac{3x-1}{5}\) - \(\frac{x}{15}\) =3

17. 2x – 3 = \(\frac{3}{10}\) x (5x – 10)

18. \(\frac{2y – 1}{3}\) - \(\frac{y - 2}{4}\) = 1

19. \(\frac{x-2}{9}\) + \(\frac{1}{3}\) = x - \(\frac{2x - 1}{3}\)

20. \(\frac{2x - 1}{3}\) - \(\frac{2x - 2}{5}\) = \(\frac{1}{3}\)

21. \(\frac{y + 7}{15}\) = 1 + \(\frac{3y - 2}{5}\)

22. \(\frac{2x - 18}{7}\) + \(\frac{x}{3}\) = 2

23. \(\frac{2x - 3}{5}\) + \(\frac{x + 3}{4}\) = \(\frac{4x + 1}{4}\)

24. \(\frac{21x - 3}{4}\) – (2x - \(\frac{1 - x}{2}\)) = x + \(\frac{5}{2}\)

25. \(\frac{x + 2}{6}\) – (\(\frac{11 - x}{3}\)- \(\frac{1}{4}\)) = \(\frac{3x - 4}{9}\)

26. \(\frac{9x + 7}{2}\) – (x - \(\frac{x - 2}{12}\) = 24

27. 0.5x +\(\frac{x}{3}\) = 0.25x + 6

28. 0.18(5x – 4) = 0.5x + 0.6

29. 2.4(3 – x) – 0.3(2x – 3) = 0

30.  0.5x – (0.6 – 0.2x) = 0.2 – 0.3x

31. \(\frac{x + 2}{x - 2}\) = \(\frac{7}{2}\)

32. \(\frac{2x + 5}{3x + 4}\)  = 5.

II. Solve the following word problems:

1. Ten added to thrice a whole number gives 40. Find the number.

2. Four-fifths of a number is greater than three-fourths of a number by 8. Find the number.

3. The sum of two consecutive multiples of 4 is 60. Find the number.

I.  1. x = 3

3. x = -3/2

4. x = -2/5

7. x = 17/36

8. x = 25/4

10. x = 11/7

12.m = 20/7

13. x = -3/13

14. t = -17

21. y = -1/4

22. x = 96/13

23. x = -2/7

24. x = 11/7

25. x = 113/6

26. x = 248/43

27. x = 72/7

28. x = 3.3

29. x = 2.1

30. x = 0.8

31. x = 18/5

32. x = -15/13

II.  1. x = 10

3. 1 st number = 28

2 nd number = 32

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Problems on Linear Equations in One Variable | Linear Equations in One Variable Practice Questions

Solved questions on linear equations in one variable are provided below with a detailed explanation. You have to eliminate brackets and expand the given linear equation. Find the variable and translate the problem to the mathematical statement. Use the conditions and solve the equation. After that verify whether the answer satisfies the condition or not.

How To Solve Linear Equation in One Variable?

Go through the step by step procedure listed for Solving Linear Equations in One Variable. They are mentioned below

• Read the given question twice or many times until you observe the parameters like what is given and what you need to find.
• To make a strategy, represent the unknown values as variables.
• Convert the word problem into mathematics.
• Portray the problem as a linear equation in one variable using the conditions provided.
• Solve the equation for unknown.
• Verify to be sure whether the answer satisfies the conditions of the problem.

Linear Equations in One Variable Word Problems with Solutions

The sum of two numbers is 44. If one exceeds the other by 6, find the numbers?

Let x be one of the two numbers.

Then, another number is (x + 6).

The Sum of two numbers is 44.

x + x + 6 = 44

2x + 6 = 44

2x = 44 – 6

The second number is x + 6 = 19 + 6 = 25

So, the two numbers are 19 and 25.

The sum of four consecutive multiples of 5 is 650. Find these multiples?

If x is a multiple of 5, the next multiple is x + 5, third number is x + 10, and fourth multiple is x + 15.

The sum of four consecutive multiples of 5 is 650

x + x + 5 + x + 10 + x + 15 = 650

4x + 30 = 650

Subtract 30 from both sides

4x + 30 – 30 = 650 – 30

Divide both sides by 4.

4x/4 = 620/4

The first multiple is 155, the second multiple is 155 + 5 = 160, the third multiple is 155 + 10 = 165, fourth multiple is 155 + 15 = 170.

So, the consecutive multiple of 5 is 155, 160, 165, and 170.

Two numbers are in the ratio 8 : 1. If they differ by 126, what are those numbers?

Given ratio is 8:1

From the ratio two numbers can be assumed as 8x, x.

The difference between numbers is 126

8x – x = 126

Divide both sides by 7.

7x/7 = 126/7

8x = 8 x 18 = 144.

Hence, two numbers are 18, 144.

The ratio of the three numbers is 5: 6: 7. If the sum of those three numbers is 54, find those numbers?

The given ratio of three numbers are 5: 6: 7

From the ratio, the numbers can be assumed as 5x, 6x, and 7x.

The sum of three numbers is 54

5x + 6x + 7x = 54

Divide both sides by 18.

18x / 18 = 54 / 18

So, the numbers be

5x = 5 x 3 = 15

6x = 6 x 3 = 18

7x = 7 x 3 = 21.

Hence, the three numbers are 15, 18, and 21.

If you subtract 2/3 from a number and multiply the result by 2/3, you will obtain 1/6. What is the number?

Let us say x is the required number.

From the given information, we can write

(x – 2/3) * 2/3 = 1/6

2x/3 – 4/9 = 1/6

2x/3 = 1/6 + 4/9

2x/3 = (3 + 8)/18

2x/3 = 11/18

Cross multiply the fractions.

2x * 18 = 11 * 3

So, the number is 33/36.

A total of $1500 is distributed among 100 persons as compensation for the work. The given compensation is either of $50 or $100. Find the number of compensations of each type?

Total number of compensations = 100

Let the number of compensations of $50 is x

Then the number of compensations of $100 is ($100 – x)

Amount spend on x compensations of $50 = $50x

Amount spend on (150 – x) compensations of $100 = $100(100 – x)

The total amount spent for compensation = $1500

According to the question,

50x + 100(100 – x) = 1500

50x + 10000 – 100x = 1500

-50x + 10000 = 1500

-50x = 1500 – 10000

-50x = -8500

x = 8500 / 50

100 – x = 100 – 170 = 70

Therefore, compensation of $50 are 170, and compensations of $100 are 70.

Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.

Let one part be x.

Then other art = 28 – x

It is given 6/5 of one part = 2/3 of the other.

6x/5 = 2/3(28 – x)

6x/5 = (28 x 2)/3 – 2x/3

6x/5 + 2x/3 = 56/3

(18x + 10x) / 15 = 56/3

28x/15 = 56/3

28x * 3 = 56 * 15

x = 840 / 84

Then the two parts are 10 and 28 – 10 = 18.

The numerator of a rational number is less than its denominator by 3. If the denominator is increased by 7 and the numerator is decreased by 1, the new number becomes 3/2. Find the original number?

Let the denominator of a rational number = x

Then the numerator of the rational number = x – 3

When denominator is increased by 7, then new denominator = x + 7

When the numerator is decreased by 1, then new numerator = x – 3 – 1 = x – 4

The new number formed = 3/2

(x – 4) / (x + 7) = 3/2

Cross multiply.

2(x – 4) = 3(x + 7)

2x – 8 = 3x + 21

-8 – 21 = 3x – 2x

The original number i.e., (x – 3) / x = (-29 – 3) / -29 = (-32)/-29 = 32/29.

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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 ‍