Linear Equations in One and Two Variables Practice Questions — GMAT

Question 1Question →

A manufacturing shop produces two models of custom bicycle frames: Standard and Deluxe. Producing each Standard frame requires $2$ hours of welding and $1$ hour of painting. Producing each Deluxe frame requires $3$ hours of welding and $2$ hours of painting. During a single week, the shop logged a total of $130$ hours of welding and $75$ hours of painting for these two models. How many Deluxe bicycle frames were produced during that week?

Show answer & explanation

Answer: 20

Answer

The number of Deluxe bicycle frames produced during that week is 20.

Translating the resource limitations into a system of two linear equations yields

2x + 3y = 130

for welding hours and

x + 2y = 75

for painting hours, where

x

and

y

represent the number of Standard and Deluxe frames respectively. Expressing

x

in terms of

y

from the painting equation gives

x = 75 - 2y

. Substituting this expression into the welding equation gives

2(75 - 2y) + 3y = 130

, which simplifies to

150 - y = 130

, giving

y = 20

Deluxe frames.

Step-by-Step Solution

1

Define variables for the unknowns.

Let

x

be the number of Standard bicycle frames produced and

y

be the number of Deluxe bicycle frames produced.

Assigning variables allows us to translate the problem into algebraic expressions.

2

Set up a system of linear equations.

Welding constraint:

2x + 3y = 130

Painting constraint:

x + 2y = 75

Each constraint represents the sum of hours spent on Standard and Deluxe frames for that process.

3

Solve the system using elimination or substitution.

Multiply the painting equation by

2

:

2x + 4y = 150

.
Subtract the welding equation (

2x + 3y = 130

) from this result:

(2x + 4y) - (2x + 3y) = 150 - 130 \Rightarrow y = 20

.

Eliminating

x

directly solves for

y

, which is the requested quantity (Deluxe frames).

Key Concept

Solving systems of two linear equations with two variables by substitution or elimination.

Estimated Time:1m 30s

Question 2Question →

What is the sum of all real solutions to the absolute value equation $|2x - 3| + |x + 5| = 12 - x$ ?

-\frac{9}{2}

\frac{5}{2}

-\frac{1}{2}

\frac{9}{2}

\frac{19}{2}

Show answer & explanation

Answer: $-\frac{9}{2}$

Answer

The sum of all real solutions to the equation is

-\frac{9}{2}

.

By splitting the real number line into three intervals based on the critical points

x = -5

and

x = \frac{3}{2}

, we find two valid solutions:

x = \frac{5}{2}

(from

x \ge \frac{3}{2}

) and

x = -7

(from

x < -5

). Their sum is

\frac{5}{2} + (-7) = -\frac{9}{2}

.

Step-by-Step Solution

1

Identify critical points for the absolute value expressions

The critical points are

x = \frac{3}{2}

and

x = -5

.

The expressions inside the absolute values,

2x - 3

and

x + 5

, change signs at

x = \frac{3}{2}

and

x = -5

respectively.

2

Evaluate Region 1 (

x \ge \frac{3}{2}

)

(2x - 3) + (x + 5) = 12 - x \implies 3x + 2 = 12 - x \implies 4x = 10 \implies x = \frac{5}{2}

.

In this region, both

2x - 3 \ge 0

and

x + 5 > 0

, so absolute value bars can be removed directly. Since

\frac{5}{2} \ge \frac{3}{2}

,

x = \frac{5}{2}

is a valid solution.

3

Evaluate Region 2 (

-5 \le x < \frac{3}{2}

)

-(2x - 3) + (x + 5) = 12 - x \implies -x + 8 = 12 - x \implies 8 = 12

(No solution).

In this interval,

2x - 3 < 0

while

x + 5 \ge 0

. The resulting equation produces a contradiction, so there are no solutions in this interval.

4

Evaluate Region 3 (

x < -5

)

-(2x - 3) - (x + 5) = 12 - x \implies -3x - 2 = 12 - x \implies -2x = 14 \implies x = -7

.

In this region, both

2x - 3 < 0

and

x + 5 < 0

. Since

-7 < -5

,

x = -7

is a valid solution.

5

Calculate the sum of all valid solutions

\frac{5}{2} + (-7) = \frac{5}{2} - \frac{14}{2} = -\frac{9}{2}

.

Combining the valid roots from Region 1 and Region 3 gives the final requested sum.

Key Concept

Solving piecewise linear equations involving multiple absolute value terms.

Estimated Time:2m 30s

Question 3Question →

A courier service calculates its total delivery charge using a fixed base fee plus a constant per-mile rate. On Monday, a delivery of 15 miles received a 20% discount on the fixed base fee and a 25% surcharge on the per-mile rate, resulting in a total charge of $30.50. On Tuesday, a delivery of 20 miles incurred a 40% increase on the fixed base fee and received a 15% discount on the per-mile rate, resulting in a total charge of$ 34.40. What is the standard total delivery charge, in dollars, for a 25-mile delivery with no fee adjustments or rate changes?

Show answer & explanation

Answer: 40

Answer

The standard total delivery charge for a 25-mile delivery is 40 dollars.

Translating the scenario into linear equations gives

0.80B + 18.75r = 30.50

and

1.40B + 17.00r = 34.40

. Solving this linear system yields a standard base fee of

B = 10.00

dollars and a standard per-mile rate of

r = 1.20

dollars. Substituting these into the standard 25-mile cost expression

B + 25r

yields

10.00 + 25(1.20) = 40.00

dollars.

Step-by-Step Solution

1

Formulate linear equations from the word problem context.

System of equations:

0.80B + 18.75r = 30.50

and

1.40B + 17.00r = 34.40

.

Applying the percentage adjustments to the fixed base fee

B

and the rate per mile

r

for the given distances yields exact linear expressions.

2

Eliminate variable

B

to solve for

r

.

Multiply equations to equate coefficients of

B

:

5.60B + 131.25r = 213.50

and

5.60B + 68.00r = 137.60

. Subtracting gives

63.25r = 75.90

, so

r = 1.20

.

Finding the per-mile rate

r

allows determination of the standard mileage component.

3

Solve for base fee

B

using

r = 1.20

.

1.40B + 17(1.20) = 34.40 \implies 1.40B = 14.00 \implies B = 10.00

.

Substituting

r

into either linear equation gives the fixed base fee.

4

Calculate the target standard cost for 25 miles.

B + 25r = 10.00 + 25(1.20) = 40.00

.

Evaluating the standard pricing expression

B + 25r

with

B = 10

and

r = 1.20

gives the total cost.

Key Concept

Solving Systems of Two-Variable Linear Equations from Word Problems

Question 4Question →

Consider the system of linear equations in

x

and

y

:

kx + 6y = 18

(k + 1)x + 9y = 27

where

k

is a constant. If this system has infinitely many solutions

(x, y)

, what is the sum of all real values of

x

that satisfy the equation

|kx - 10| = 3x - 4

?

\frac{14}{5}

-\frac{16}{5}

-6

10

\frac{6}{5}

Show answer & explanation

Answer: $\frac{14}{5}$

Answer

The sum of all real values of

x

satisfying the equation is

\frac{14}{5}

(or

2.8

).

The system of linear equations has infinitely many solutions when the ratio of coefficients is constant:

\frac{k}{k+1} = \frac{6}{9} = \frac{18}{27} = \frac{2}{3}

, which yields

k = 2

. Substituting

k = 2

gives

|2x - 10| = 3x - 4

. Breaking this into cases yields

2x - 10 = 3x - 4 \implies x = -6

and

-(2x - 10) = 3x - 4 \implies x = \frac{14}{5}

. Testing

x = -6

in the original equation gives

|-22| = -22

, which is false, making

x = -6

an extraneous solution. Testing

x = \frac{14}{5}

gives

|-4.4| = 4.4

, which is true. Thus, the only real solution is

\frac{14}{5}

, making the option equal to

\frac{14}{5}

correct.

Step-by-Step Solution

1

Determine the value of the constant

k

from the system of linear equations.

k = 2

For a system of two linear equations in two variables to have infinitely many solutions, the ratio of coefficients of

x

,

y

, and the constant terms must all be equal:

\frac{k}{k+1} = \frac{6}{9} = \frac{18}{27}

. Since

\frac{6}{9} = \frac{2}{3}

, setting

\frac{k}{k+1} = \frac{2}{3}

gives

3k = 2k + 2 \implies k = 2

.

2

Substitute

k = 2

into the absolute value equation.

|2x - 10| = 3x - 4

Replacing

k

with

2

sets up the single-variable linear absolute value equation to solve.

3

Solve Case 1 where the expression inside the absolute value is non-negative (

2x - 10 \ge 0

).

x = -6

(Extraneous)

Setting

2x - 10 = 3x - 4

yields

x = -6

. However, substituting

x = -6

back into the right side gives

3(-6) - 4 = -22

. Since an absolute value cannot equal a negative number,

x = -6

is an extraneous solution.

4

Solve Case 2 where the expression inside the absolute value is negative (

2x - 10 < 0

).

x = \frac{14}{5}

(Valid)

Setting

-(2x - 10) = 3x - 4

gives

-2x + 10 = 3x - 4 \implies 5x = 14 \implies x = \frac{14}{5}

. Substituting

x = 2.8

into the right side yields

3(2.8) - 4 = 4.4 \ge 0

, and

|2(2.8) - 10| = |-4.4| = 4.4

. Thus,

x = \frac{14}{5}

is valid.

5

Sum all valid real solutions.

\frac{14}{5}

Since

x = \frac{14}{5}

is the only valid solution, the sum of all solutions is

\frac{14}{5}

.

Key Concept

Systems of linear equations with infinitely many solutions require identical proportional coefficients. Absolute value equations of the form

|A| = B

require checking candidate solutions in

B \ge 0

to eliminate extraneous roots.

Estimated Time:3m 0s

Question 5Question →

A technology manufacturing company produces two server models: Model Alpha and Model Beta. The manufacturing cost of one Model Alpha server is $400 more than twice the manufacturing cost of one Model Beta server. Last month, the company produced 15 Model Alpha servers and 40 Model Beta servers, incurring a total manufacturing cost of$ 76,000. What is the manufacturing cost, in dollars, of one Model Alpha server?

Show answer & explanation

Answer: 2400

Answer

The manufacturing cost of one Model Alpha server is $2,400.

By defining the costs as

A = 2B + 400

and setting up the total cost equation

15A + 40B = 76,000

, substituting

A

yields

70B = 70,000

, giving

B = 1,000

. Substituting

B = 1,000

back into

A = 2B + 400

gives

A = 2,400

.

Step-by-Step Solution

1

Define variables and construct the linear relation equation.

A = 2B + 400

The cost of one Model Alpha server (

A

) is

400 more than twice the cost of Model Beta (

B$).

2

Formulate the total production cost equation.

15A + 40B = 76,000

15 units of Model Alpha and 40 units of Model Beta sum to a total cost of $76,000.

3

Substitute

A

into the total cost equation and solve for

B

.

15(2B + 400) + 40B = 76,000 \implies 70B + 6,000 = 76,000 \implies B = 1,000

Substituting reduces the system to a single linear equation in one variable.

4

Calculate the value of

A

.

A = 2(1,000) + 400 = 2,400

Substitute

B = 1,000

back into the equation for

A

.

Key Concept

Solving Systems of Linear Equations by Substitution

Question 6Question →

If $x$ is a real number that satisfies the equation $|x + 7| = 2x + 3$ , what is the sum of all valid real solutions for $x$ ?

\frac{2}{3}

4

-\frac{10}{3}

-\frac{2}{3}

\frac{22}{3}

Show answer & explanation

Answer: $4$

Answer

The sum of all valid real solutions is 4.

To solve

|x + 7| = 2x + 3

, we break the absolute value into two linear cases:

x + 7 = 2x + 3

giving

x = 4

, and

x + 7 = -(2x + 3)

giving

x = -\frac{10}{3}

. We must test both candidate solutions in the original equation because absolute value outputs must be non-negative. Testing

x = 4

gives

|11| = 11

, which is true. Testing

x = -\frac{10}{3}

gives

|\frac{11}{3}| = -\frac{11}{3}

, which is false because an absolute value cannot equal a negative number. Thus,

x = 4

is the only valid solution, making the sum of all valid solutions equal to

4

.

Step-by-Step Solution

1

Set up the two linear cases for the absolute value equation

|x + 7| = 2x + 3

.

Case 1:

x + 7 = 2x + 3

; Case 2:

x + 7 = -(2x + 3)

.

By definition,

|A| = B

implies

A = B

or

A = -B

, provided

B \ge 0

.

2

Solve Case 1 for

x

.

x + 7 = 2x + 3 \implies 2x - x = 7 - 3 \implies x = 4

.

Isolating

x

gives the first candidate solution.

3

Solve Case 2 for

x

.

x + 7 = -2x - 3 \implies 3x = -10 \implies x = -\frac{10}{3}

.

Expanding the negative sign and isolating

x

gives the second candidate solution.

4

Check candidate solutions in the original equation

|x + 7| = 2x + 3

to eliminate extraneous solutions.

For

x = 4

:

|4 + 7| = 11

and

2(4) + 3 = 11

(Valid).
For

x = -\frac{10}{3}

:

|-\frac{10}{3} + 7| = \frac{11}{3}

, but

2(-\frac{10}{3}) + 3 = -\frac{11}{3}

(Extraneous, since

\frac{11}{3} \neq -\frac{11}{3}

).

An absolute value expression cannot equal a negative number, so candidate values resulting in a negative right side are invalid.

5

Calculate the sum of all valid real solutions.

Sum =

4

.

Since

x = 4

is the only valid solution, the sum is simply

4

.

Key Concept

Solving Absolute Value Linear Equations and Validating against Extraneous Solutions

Estimated Time:2m 0s

Question 7Question →

A logistics company offers two freight pricing models for oversized cargo. Model X charges a fixed monthly account fee of $\$8$ plus $\$0.60$ per kilometer traveled. Model Y charges a fixed monthly account fee of $\$54$ plus $\$0.20$ per kilometer traveled. For how many kilometers in a month will the total monthly charge under Model X be exactly $20\%$ less than the total monthly charge under Model Y?

Show answer & explanation

Answer: 80

Answer

80 kilometers

The total monthly cost under Model X is

C_X = 8 + 0.60k

and under Model Y is

C_Y = 54 + 0.20k

. The condition that Model X is 20% less than Model Y means

C_X = 0.80 C_Y

. Substituting the expressions gives

8 + 0.60k = 0.80(54 + 0.20k) = 43.2 + 0.16k

. Subtracting

0.16k

and

8

from both sides yields

0.44k = 35.2

, which simplifies to

k = 80

.

Step-by-Step Solution

1

Define variables and establish linear cost equations for both models.

Model X cost:

C_X = 8 + 0.60k

; Model Y cost:

C_Y = 54 + 0.20k

, where

k

is kilometers traveled.

Linear modeling translates flat fees and variable rates into algebraic expressions.

2

Formulate the linear equation based on the condition that Model X is 20% less than Model Y.

C_X = 0.80 C_Y \implies 8 + 0.60k = 0.80(54 + 0.20k)

.

Being 20% less than a base value means taking 80% (or 0.80) of that value.

3

Expand and simplify the algebraic equation.

8 + 0.60k = 43.2 + 0.16k \implies 0.44k = 35.2

.

Distributing 0.80 across

(54 + 0.20k)

yields

43.2 + 0.16k

, and subtracting

0.16k

and

8

isolates

k

on one side.

4

Calculate the value of

k

.

k = \frac{35.2}{0.44} = 80

.

Dividing

35.2

by

0.44

gives the exact number of kilometers required.

Key Concept

Linear Equations in One and Two Variables

Question 8Question →

An investor holds $x$ shares of Stock X and $y$ shares of Stock Y in a financial portfolio. The total monetary value of these holdings is $\$6,300$ , modeled by the linear equation $40x + 20y = 6,300$ . If the ratio of the number of shares of Stock X to the total number of shares held is $2$ to $5$ , what is the total number of shares of Stock X and Stock Y in the portfolio?

90

135

210

225

245

Show answer & explanation

Answer: 225

Answer

225

The correct answer is 225. Expressing the part-to-whole ratio

\frac{x}{x+y} = \frac{2}{5}

algebraically gives

3x = 2y

, or

y = 1.5x

. Substituting

1.5x

for

y

in the value equation

40x + 20y = 6,300

yields

70x = 6,300

, so

x = 90

. Then

y = 135

, and the total number of shares is

90 + 135 = 225

.

Step-by-Step Solution

1

Set up the linear relationship between

x

and

y

using the given ratio.

\frac{x}{x + y} = \frac{2}{5} \implies 5x = 2x + 2y \implies 3x = 2y \implies y = 1.5x

The problem specifies that the ratio of Stock X shares (

x

) to total shares (

x + y

) is

2:5

.

2

Substitute

y = 1.5x

into the portfolio value equation

40x + 20y = 6,300

.

40x + 20(1.5x) = 6,300 \implies 40x + 30x = 6,300 \implies 70x = 6,300 \implies x = 90

Replacing

y

with an equivalent expression in terms of

x

reduces the system to a single linear equation in one variable.

3

Calculate

y

and determine the total number of shares

x + y

.

y = 1.5(90) = 135 \implies x + y = 90 + 135 = 225

The question asks for the total combined number of shares of both stocks held in the portfolio.

Key Concept

Solving systems of linear equations formed by combining a linear value equation with a ratio relationship.

Question 9Question →

If $x$ is a real number that satisfies the equation $|2x - 1| = 3x + 11$ , what is the value of $x^2 + 2x$ ?

0

3

8

35

120

Show answer & explanation

Answer: $0$

Answer

The value of

x^2 + 2x

is

0

.

Solving the equation

|2x - 1| = 3x + 11

produces two algebraic candidates:

x = -12

and

x = -2

. Substituting

x = -12

into the right side yields

3(-12) + 11 = -25

. Since absolute value expressions cannot be negative,

x = -12

is an extraneous solution. Substituting

x = -2

yields

|-5| = 5

, which is valid. Evaluating

x^2 + 2x

at

x = -2

gives

(-2)^2 + 2(-2) = 0

.

Step-by-Step Solution

1

Set up the two linear cases for the absolute value equation

|2x - 1| = 3x + 11

.

Case 1:

2x - 1 = 3x + 11

; Case 2:

2x - 1 = -(3x + 11)

.

By definition,

|u| = c

implies

u = c

or

u = -c

(provided

c \geq 0

).

2

Solve each linear equation for candidate values of

x

.

From Case 1:

x = -12

. From Case 2:

2x - 1 = -3x - 11 \implies 5x = -10 \implies x = -2

.

Isolate

x

algebraically in both equations.

3

Check candidate solutions in the original equation to eliminate extraneous roots.

For

x = -12

:

|2(-12) - 1| = |-25| = 25

, but

3(-12) + 11 = -25 \neq 25

(extraneous). For

x = -2

:

|2(-2) - 1| = |-5| = 5

, and

3(-2) + 11 = 5

(valid solution).

An absolute value cannot equal a negative number; substituting back is mandatory.

4

Evaluate the target expression

x^2 + 2x

using the valid root

x = -2

.

(-2)^2 + 2(-2) = 4 - 4 = 0

.

Substitute the verified real solution into the given expression.

Key Concept

Solving Absolute Value Linear Equations and Checking for Extraneous Solutions

Estimated Time:1m 30s

Question 10Question →

A boutique bakery sells custom gift baskets containing two types of pastries: almond tarts and chocolate croissants. Basket A contains 4 almond tarts and 3 chocolate croissants and costs $62. Basket B contains 3 almond tarts and 4 chocolate croissants and costs$ 57. What is the combined cost of 1 almond tart and 1 chocolate croissant?

$15

$17

$19

$30

$34

Show answer & explanation

Answer: $17

Answer

The combined cost of 1 almond tart and 1 chocolate croissant is $17.

By representing the prices of an almond tart and a chocolate croissant as

t

and

c

, we form the system

4t + 3c = 62

and

3t + 4c = 57

. Adding both equations yields

7t + 7c = 119

. Dividing both sides by 7 gives

t + c = 17

, which directly provides the combined price of 1 almond tart and 1 chocolate croissant.

Step-by-Step Solution

1

Set up a system of two linear equations using variables for the prices of the pastries.

Let

t

be the price of one almond tart and

c

be the price of one chocolate croissant.
Equation 1:

4t + 3c = 62

Equation 2:

3t + 4c = 57

Translating the problem statement into standard linear algebraic equations.

2

Add the two linear equations together.

(4t + 3c) + (3t + 4c) = 62 + 57 \implies 7t + 7c = 119

Symmetric coefficients allow finding the sum of

t + c

without needing to solve for individual variables first.

3

Divide the combined equation by 7 to solve for

(t + c)

.

t + c = \frac{119}{7} = 17

Factoring out 7 gives

7(t + c) = 119

, which simplifies directly to the requested sum.

Key Concept

Solving Systems of Linear Equations via Symmetric Coefficient Addition

Alternative Method

Alternatively, solve for one variable first: multiply Equation 1 by 3 (

12t + 9c = 186

) and Equation 2 by 4 (

12t + 16c = 228

). Subtracting the equations gives

7c = 42 \implies c = 6

. Substituting

c = 6

into Equation 1 gives

4t + 18 = 62 \implies 4t = 44 \implies t = 11

. Thus,

t + c = 11 + 6 = 17

.

Estimated Time:1m 15s

Question 11Question →

If real numbers

x

and

y

satisfy the system of equations:

3x - 5y = 28

x(x - 2y) = 3y(x - 2y)

where

x \neq 2y

, what is the value of

x + y

?

7

21

28

35

84

Show answer & explanation

Answer: 28

Answer

The value of

x + y

is 28.

Because

x \neq 2y

, the expression

(x - 2y)

is non-zero. Dividing both sides of the second equation

x(x - 2y) = 3y(x - 2y)

by

(x - 2y)

gives

x = 3y

. Substituting

x = 3y

into the first equation

3x - 5y = 28

yields

3(3y) - 5y = 28

, which simplifies to

4y = 28

, so

y = 7

. Then

x = 3(7) = 21

. The required sum

x + y

is

21 + 7 = 28

.

Step-by-Step Solution

1

Simplify the second equation using the given constraint.

Since

x \neq 2y

, the factor

(x - 2y) \neq 0

. Dividing both sides of

x(x - 2y) = 3y(x - 2y)

by

(x - 2y)

yields

x = 3y

.

Dividing by a non-zero expression is algebraically valid and simplifies the second equation into a linear relationship.

2

Substitute

x = 3y

into the first linear equation.

3(3y) - 5y = 28 \implies 9y - 5y = 28 \implies 4y = 28 \implies y = 7

.

Substitution reduces the system to a single linear equation in one variable.

3

Calculate

x

and the required sum

x + y

.

x = 3(7) = 21

, so

x + y = 21 + 7 = 28

.

Demonstrates full solution of the target expression.

Key Concept

Solving Systems of Linear Equations by Algebraic Reduction

Question 12Question →

A logistics company models its total daily delivery cost $C$ , in dollars, as a linear function of the total distance driven $d$ , in miles, given by $C = m \cdot d + b$ , where $m$ and $b$ are constants. The total daily cost increases by $\$180$ when the distance driven increases by $120$ miles. If the total daily delivery cost is $\$450$ when $250$ miles are driven, what is the total daily delivery cost when $400$ miles are driven?

$600

$630

$675

$720

$810

Show answer & explanation

Answer: $675

Answer

The total daily delivery cost when 400 miles are driven is $675.

The cost function follows the linear form

C = m \cdot d + b

. The marginal rate

m = \frac{\Delta C}{\Delta d} = \frac{180}{120} = 1.5

dollars per mile. Substituting

d = 250

and

C = 450

gives

450 = 1.5(250) + b

, which yields

b = 75

. Substituting

d = 400

into

C = 1.5d + 75

produces

C = 600 + 75 = 675

dollars.

Step-by-Step Solution

1

Determine the slope (variable cost per mile)

m

from the rate of change.

m = \frac{\Delta C}{\Delta d} = \frac{180}{120} = 1.5

dollars per mile.

The slope of a linear equation represents the constant rate of change between the dependent and independent variables.

2

Calculate the fixed base cost

b

using the given data point

(d = 250, C = 450)

.

450 = 1.5(250) + b \implies 450 = 375 + b \implies b = 75

.

Substituting a known point into the linear model determines the vertical intercept (fixed cost).

3

Evaluate the linear function for

d = 400

miles.

C = 1.5(400) + 75 = 600 + 75 = 675

dollars.

Applying the fully determined linear equation

C = 1.5d + 75

yields the total cost for the required distance.

Key Concept

Linear Modeling in Two Variables (

y = mx + b

)

Question 13Question →

If $|x - 5| = 2x - 4$ , what is the value of $x$ ?

3

-1

-1 and 3

1

5

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

Solving the absolute value equation yields two candidate values,

x = 3

and

x = -1

. Substituting

x = 3

into the original equation gives

|3 - 5| = 2

and

2(3) - 4 = 2

, which makes the statement true. Therefore,

x = 3

is the only valid solution.

Step-by-Step Solution

1

Set up the two cases for the absolute value equation

|x - 5| = 2x - 4

.

Case 1:

x - 5 = 2x - 4

; Case 2:

x - 5 = -(2x - 4)

.

By definition,

|a| = b

implies

a = b

or

a = -b

when

b \ge 0

.

2

Solve Case 1:

x - 5 = 2x - 4

.

x = -1

.

Subtract

x

and add

4

to both sides.

3

Solve Case 2:

x - 5 = -2x + 4

.

3x = 9 \implies x = 3

.

Add

2x

and add

5

to both sides.

4

Test potential solutions in the original equation to eliminate extraneous roots.

For

x = -1

:

|-1 - 5| = 6 \neq 2(-1) - 4 = -6

(Invalid). For

x = 3

:

|3 - 5| = 2 = 2(3) - 4 = 2

(Valid).

The absolute value of a real number cannot be negative, so the right-hand side

2x - 4

must be non-negative.

Key Concept

Absolute Value Equations and Extraneous Solutions

Question 14Question →

If $x$ satisfies the equation $|8x + 13| = 10x + 5$ , what is the sum of all valid real solutions for $x$ ?

4

3

-1

-4

5

Show answer & explanation

Answer: 4

Answer

The sum of all valid real solutions for

x

is 4.

The solution

x = 4

is the only value of

x

that satisfies

|8x + 13| = 10x + 5

because evaluating both sides yields

45 = 45

. The alternative algebraic candidate,

x = -1

, produces a right-hand side of

-5

, which is impossible for an absolute value expression. Thus, the sum of all valid real solutions is simply 4.

Step-by-Step Solution

1

Split the absolute value equation into two linear case equations

Case 1:

8x + 13 = 10x + 5

; Case 2:

8x + 13 = -(10x + 5)

By definition of absolute value,

|u| = v

implies

u = v

or

u = -v

, provided

v \ge 0

.

2

Solve Case 1 for

x

2x = 8 \implies x = 4

Subtract

8x

and

5

from both sides to isolate

x

.

3

Solve Case 2 for

x

8x + 13 = -10x - 5 \implies 18x = -18 \implies x = -1

Expand the right-hand side and combine like terms to isolate

x

.

4

Test candidate solutions in the original equation to eliminate extraneous roots

For

x = 4

:

|8(4)+13| = |45| = 45

and

10(4)+5 = 45

(Valid). For

x = -1

:

|8(-1)+13| = |5| = 5

and

10(-1)+5 = -5

(Extraneous).

An absolute value output cannot equal a negative number.

5

Calculate the sum of all valid solutions

The only valid solution is

x = 4

, so the sum is

4

.

Only valid roots must be summed.

Key Concept

Absolute Value Equations and Checking for Extraneous Solutions

Question 15Question →

If real numbers $x$ and $y$ satisfy the system of linear equations:

7x + 4y = 113

3x + 6y = 87

what is the value of $x + y$ ?

Show answer & explanation

Answer: 20

Answer

The value of

x + y

is 20.

Adding the two equations yields

10x + 10y = 200

. Dividing the entire equation by 10 directly gives

x + y = 20

. Alternatively, solving the system yields

x = 11

and

y = 9

, whose sum is

11 + 9 = 20

.

Step-by-Step Solution

1

Add the two equations together to combine like terms.

(7x + 4y) + (3x + 6y) = 113 + 87 \implies 10x + 10y = 200

Recognizing that adding the equations creates symmetric coefficients of 10 for both

x

and

y

allows a direct solution for the sum

(x + y)

.

2

Divide the combined equation by 10.

\frac{10(x + y)}{10} = \frac{200}{10} \implies x + y = 20

Isolating

(x + y)

directly avoids the extra computation of solving for

x

and

y

individually.

Key Concept

Solving Systems of Linear Equations by Combination

Estimated Time:1m 30s

Question 16Question →

A car rental agency charges a fixed daily rate plus a constant fee per mile driven. A customer who rented a car for 3 days and drove 150 miles paid a total of $180. Another customer who rented the same model of car for 5 days and drove 200 miles paid a total of$ 270. What is the fixed daily rate, in dollars, charged by the rental agency?

Show answer & explanation

Answer: 30

Answer

The fixed daily rate charged by the rental agency is 30 dollars.

Formulating the equations

3d + 150m = 180

and

5d + 200m = 270

and solving for the daily rate

d

yields

d = 30

.

Step-by-Step Solution

1

Define variables and formulate the system of linear equations.

Let

d

be the fixed daily rate (in dollars) and

m

be the cost per mile driven (in dollars). The scenario translates to:
Equation 1:

3d + 150m = 180

Equation 2:

5d + 200m = 270

Total charge is the linear combination of daily fixed costs and per-mile variable costs.

2

Simplify Equation 1 to express

d

in terms of

m

.

Dividing Equation 1 by 3 yields

d + 50m = 60

, so

d = 60 - 50m

.

Simplifying equations reduces computation error when using substitution.

3

Substitute the expression for

d

into Equation 2 to solve for

m

.

5(60 - 50m) + 200m = 270 \Rightarrow 300 - 250m + 200m = 270 \Rightarrow -50m = -30 \Rightarrow m = 0.60

Substitution eliminates variable

d

, leaving a linear equation in one variable.

4

Calculate the fixed daily rate

d

.

d = 60 - 50(0.60) = 60 - 30 = 30

Substituting

m = 0.60

gives the exact daily fixed rate.

Key Concept

Setting up and solving a system of two linear equations in two variables using elimination or substitution.

Question 17Question →

If $3(x - 2) + 4 = 2x + 11$ , what is the value of $x$ ?

Show answer & explanation

Answer: 13

Answer

The value of

x

is 13.

Expanding the left side yields

3x - 6 + 4 = 3x - 2

. Equating this to the right side gives

3x - 2 = 2x + 11

. Subtracting

2x

from both sides gives

x - 2 = 11

, and adding

2

to both sides yields

x = 13

.

Step-by-Step Solution

1

Expand the left side of the equation

3x - 6 + 4 = 2x + 11

Apply the distributive property

3(x - 2) = 3x - 6

.

2

Combine like constant terms on the left side

3x - 2 = 2x + 11

Combine

-6

and

+4

to get

-2

.

3

Subtract

2x

from both sides

x - 2 = 11

Move variable terms to one side of the equation.

4

Add 2 to both sides

x = 13

Isolate the variable

x

.

Key Concept

Solving Linear Equations in One Variable

Question 18Question →

If $|2x - 3| = 7$ , which of the following is a possible value of $x$ ?

5

-5

2

10

-4

Show answer & explanation

Answer: $5$

Answer

The value

5

is a valid solution to the given absolute value equation.

The absolute value equation

|2x - 3| = 7

splits into two linear equations:

2x - 3 = 7

and

2x - 3 = -7

. Solving the first equation gives

2x = 10

, so

x = 5

. Solving the second gives

2x = -4

, so

x = -2

. Among the listed options,

5

is the correct solution.

Step-by-Step Solution

1

Set up the two cases for the absolute value equation

|2x - 3| = 7

.

Case 1:

2x - 3 = 7

; Case 2:

2x - 3 = -7

.

An absolute value expression

|u| = c

(where

c \geq 0

) resolves to

u = c

or

u = -c

.

2

Solve Case 1:

2x - 3 = 7

.

2x = 10 \implies x = 5

.

Add 3 to both sides and divide by 2.

3

Solve Case 2:

2x - 3 = -7

.

2x = -4 \implies x = -2

.

Add 3 to both sides and divide by 2.

4

Compare the solutions (

x = 5

and

x = -2

) with the given choices.

The value

5

is present among the options.

Identifies the correct choice matching one of the calculated solutions.

Key Concept

Solving Linear Absolute Value Equations

Estimated Time:45s

Question 19Question →

A financial analyst models a company's total annual revenue, $R(t)$ , and total annual expenses, $E(t)$ , in thousands of dollars, as linear functions of time $t$ , where $t$ represents the number of years since 2020. In 2022 ( $t = 2$ ), the annual revenue was $\$850,000$ and annual expenses were $\$610,000$ . In 2025 ( $t = 5$ ), the annual revenue reached $\$1,120,000$ while annual expenses reached $\$700,000$ . Assuming these linear trends continue, in which calendar year will the company's annual profit (defined as total annual revenue minus total annual expenses) reach exactly $\$540,000$ ?

Show answer & explanation

Answer: 2027

Answer

The company's annual profit will reach $540,000 in the year 2027.

The linear profit function increases at a constant rate of

60,000 per year from a base profit of

240,000 in 2022 (

t = 2

). Solving

240 + 60(t - 2) = 540

yields

t = 7

, which corresponds to the calendar year 2027.

Step-by-Step Solution

1

Calculate the annual profit in thousands of dollars for the known years 2022 (

t = 2

) and 2025 (

t = 5

).

In 2022 (

t = 2

), profit

P(2) = 850 - 610 = 240

thousand dollars. In 2025 (

t = 5

), profit

P(5) = 1120 - 700 = 420

thousand dollars.

Profit is defined as revenue minus expenses.

2

Determine the linear profit function

P(t) = mt + b

.

The rate of change of profit per year is

m = \frac{420 - 240}{5 - 2} = \frac{180}{3} = 60

thousand dollars per year.

Since both revenue and expenses are linear functions of time

t

, their difference

P(t) = R(t) - E(t)

is also a linear function of

t

.

3

Set up the linear equation for profit using point-slope form.

P(t) - 240 = 60(t - 2) \implies P(t) = 60t + 120

.

Using the point

(2, 240)

and slope

m = 60

establishes the complete linear equation for annual profit.

4

Solve the linear equation for

t

when profit

P(t) = 540

thousand dollars.

60t + 120 = 540 \implies 60t = 420 \implies t = 7

.

Setting the profit equal to 540 gives the value of

t

years after 2020.

5

Convert the value of

t

into the target calendar year.

Calendar Year

= 2020 + 7 = 2027

.

Since

t

represents years elapsed since 2020,

t = 7

corresponds to calendar year 2027.

Key Concept

Linear Modeling and Linear Equations in Two Variables

Question 20Question →

A production facility uses two automated machines, Alpha and Beta, which operate at constant hourly production rates. Machine Alpha produces $x$ units per hour and Machine Beta produces $y$ units per hour. When both machines operate together for 4 hours, followed by Machine Alpha operating alone for 3 hours, the facility produces a total of 310 units. When Machine Beta operates alone for 2 hours, followed by both machines operating together for 5 hours, the facility produces a total of 325 units. If Machine Alpha's hourly rate is increased by $20\%$ and Machine Beta's hourly rate is decreased by $20\%$ , how many total units will both machines produce if they work together for 8 hours at their new rates?

Show answer & explanation

Answer: 448

Answer

The total number of units produced by both machines working together for 8 hours under the modified rates is 448.

Setting up the system of equations based on individual hours worked yields

7x + 4y = 310

and

5x + 7y = 325

. Solving this system gives original rates

x = 30

units/hr and

y = 25

units/hr. Applying the

20\%

increase to Alpha (

36

units/hr) and

20\%

decrease to Beta (

20

units/hr) gives a combined rate of

56

units/hr. Multiplying by

8

hours gives the final answer of

448

units.

Step-by-Step Solution

1

Translate the given operational scenarios into a system of two linear equations.

Equation 1:

7x + 4y = 310

; Equation 2:

5x + 7y = 325

.

Operating both machines for 4 hours and Alpha alone for 3 hours means Alpha works

4 + 3 = 7

hours while Beta works

4

hours, giving

7x + 4y = 310

. Operating Beta alone for 2 hours and both for 5 hours means Alpha works

5

hours while Beta works

2 + 5 = 7

hours, giving

5x + 7y = 325

.

2

Solve the system of linear equations for variables

x

and

y

.

x = 30

and

y = 25

.

Eliminating

y

by multiplying the first equation by 7 and the second by 4 yields

49x - 20x = 2170 - 1300 \implies 29x = 870 \implies x = 30

. Substituting

x = 30

back into

7x + 4y = 310

gives

210 + 4y = 310 \implies 4y = 100 \implies y = 25

.

3

Calculate the modified production rates after percentage adjustments.

New rate for Alpha is

36

units/hr; new rate for Beta is

20

units/hr.

A

20\%

increase on

x = 30

yields

30 \times 1.20 = 36

. A

20\%

decrease on

y = 25

yields

25 \times 0.80 = 20

.

4

Compute total combined output over 8 hours.

Total units produced =

448

.

Combined modified rate is

36 + 20 = 56

units per hour. Total production over 8 hours is

56 \times 8 = 448

.

Key Concept

Linear Equations in One and Two Variables

Estimated Time:2m 30s

Linear Equations in One and Two Variables

Answer

Step-by-Step Solution

Key Concept

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Answer

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