Linear Equations in One Variable Practice Questions — GRE General Test

Question 1Question →

If $\frac{2x - 3}{4} - \frac{x - 5}{3} = \frac{x + 15}{12}$ , what is the value of $3x - 2$ ?

10

4

37

55

130

Show answer & explanation

Answer: 10

Answer

The value of

3x - 2

is

10

.

Multiplying the given linear equation by the least common denominator 12 yields

3(2x - 3) - 4(x - 5) = x + 15

. Expanding both sides gives

6x - 9 - 4x + 20 = x + 15

, which simplifies to

2x + 11 = x + 15

, leading to

x = 4

. Substituting

x = 4

into

3x - 2

gives

3(4) - 2 = 10

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the equation by the least common multiple of 4, 3, and 12, which is 12.

12 \cdot \left(\frac{2x - 3}{4}\right) - 12 \cdot \left(\frac{x - 5}{3}\right) = 12 \cdot \left(\frac{x + 15}{12}\right) \implies 3(2x - 3) - 4(x - 5) = x + 15

Eliminating fractions simplifies the algebraic manipulation.

2

Distribute the coefficients through the terms in parentheses, paying careful attention to the negative sign on the second term.

6x - 9 - 4x + 20 = x + 15

Distributing

-4

across

(x - 5)

yields

-4x + 20

.

3

Combine like terms on the left side of the equation.

2x + 11 = x + 15

Grouping variable terms

(6x - 4x = 2x)

and constant terms

(-9 + 20 = 11)

.

4

Isolate the variable

x

by subtracting

x

and

11

from both sides.

x = 4

Solves the linear equation for

x

.

5

Substitute

x = 4

into the target expression

3x - 2

.

3(4) - 2 = 12 - 2 = 10

Calculates the final answer requested in the stem.

Key Concept

Linear Equations with Fractional Coefficients

Question 2Question →

For a real constant

k \neq -1

, consider the linear equation in one variable

x

:

\frac{2(x - 3)}{k + 1} - \frac{x + 1}{3} = 1

Which of the following values of

k

result in a solution

x

that is a positive integer? Select all such values.

Select all that apply

-2

0

1

2

3

Show answer & explanation

Answer: -2; 2; 3

Answer

The values of

k

that yield a positive integer solution for

x

are

-2

,

2

, and

3

.

Solving the linear equation for

x

in terms of

k

gives

x = \frac{4k + 22}{5 - k}

. Substituting

k = -2

yields

x = 2

, substituting

k = 2

yields

x = 10

, and substituting

k = 3

yields

x = 17

. All three resulting values of

x

are positive integers.

Step-by-Step Solution

1

Clear denominators by multiplying the entire equation by

3(k + 1)

.

6(x - 3) - (x + 1)(k + 1) = 3(k + 1)

Eliminating fractional terms simplifies isolation of the variable

x

.

2

Expand all terms on both sides of the equation.

6x - 18 - (kx + x + k + 1) = 3k + 3 \implies 5x - kx - 19 - k = 3k + 3

Carefully distribute negative signs and combine like terms.

3

Group terms containing

x

on the left side and constant/k terms on the right side.

(5 - k)x = 4k + 22 \implies x = \frac{4k + 22}{5 - k}

Express

x

explicitly as a rational function of the parameter

k

.

4

Evaluate

x

for each given option to determine which produce positive integers.

For

k = -2

:

x = 14/7 = 2

(integer);
For

k = 0

:

x = 22/5 = 4.4

(not integer);
For

k = 1

:

x = 26/4 = 6.5

(not integer);
For

k = 2

:

x = 30/3 = 10

(integer);
For

k = 3

:

x = 34/2 = 17

(integer).

Test each candidate value of

k

against the condition that

x

must be a positive integer.

Key Concept

Solving parametric linear equations in one variable and evaluating integer solutions

Estimated Time:2m 30s

Question 3Question →

In the linear equation below,

k

is a constant:

\frac{x - k}{2} + \frac{2x + 1}{5} = \frac{3x - 4}{10}

If

x

must be a positive integer, which of the following values could be

k

? Select all such values.

Select all that apply

-6

6

10

12

15

Show answer & explanation

Answer: $6$ ; $12$

Answer

The values of

k

that satisfy the condition are

6

and

12

.

Clearing denominators gives

6x = 5k - 6

, or

x = \frac{5k - 6}{6}

. For

x

to be a positive integer,

5k - 6

must be positive and divisible by

6

. This requires

k

to be a positive multiple of

6

. The options

6

and

12

satisfy both criteria, yielding

x = 4

and

x = 9

respectively.

Step-by-Step Solution

1

Clear denominators by multiplying the entire equation by the least common denominator (LCD), which is 10.

5(x - k) + 2(2x + 1) = 3x - 4

Eliminating fractions simplifies further algebraic manipulation.

2

Expand terms and combine like terms on the left side of the equation.

5x - 5k + 4x + 2 = 3x - 4 \implies 9x - 5k + 2 = 3x - 4

Distribute constants across parentheses to isolate variable expressions.

3

Isolate

x

in terms of

k

.

6x = 5k - 6 \implies x = \frac{5k - 6}{6}

Express

x

as an explicit linear function of

k

to test divisibility and positivity conditions.

4

Determine constraints on

k

for

x

to be a positive integer.

5k

must be divisible by

6

(so

k

must be a multiple of

6

) and

5k - 6 > 0 \implies k > 1.2

.

Since

5

and

6

share no common factors,

k

must be a positive multiple of

6

.

5

Evaluate the choices against the constraints.

Among the options, only

6

and

12

are positive multiples of

6

, producing

x = 4

and

x = 9

respectively.

Both values produce integer solutions greater than zero.

Key Concept

Solving single-variable linear equations with rational coefficients and applying divisibility and inequality constraints.

Question 4Question →

Consider the linear equation in one variable

x

:

\frac{2x - 3}{3} - \frac{x - 5}{4} = \frac{x + 9}{6}

If

x

is the solution to this equation, what is the value of

4x - 1

?

5

7

11

19

59

Show answer & explanation

Answer: 19

Answer

19

To solve the linear equation, first multiply all terms by the least common denominator, 12, yielding 4(2x - 3) - 3(x - 5) = 2(x + 9). Expanding both sides yields 8x - 12 - 3x + 15 = 2x + 18. Combining like terms on the left side yields 5x + 3 = 2x + 18. Subtracting 2x and 3 from both sides gives 3x = 15, so x = 5. Substituting x = 5 into the target expression 4x - 1 gives 4(5) - 1 = 19.

Step-by-Step Solution

1

Clear the fractional denominators by multiplying every term by the least common multiple (LCM) of 3, 4, and 6, which is 12.

12 \cdot \left(\frac{2x - 3}{3}\right) - 12 \cdot \left(\frac{x - 5}{4}\right) = 12 \cdot \left(\frac{x + 9}{6}\right) \implies 4(2x - 3) - 3(x - 5) = 2(x + 9)

Multiplying every term by 12 eliminates all denominators without changing the equation's solution set.

2

Distribute the factors across the binomial terms inside parentheses.

8x - 12 - 3x + 15 = 2x + 18

Distributing -3 across (x - 5) yields -3x + 15 because the product of two negative numbers is positive.

3

Combine like terms on the left side of the equation.

5x + 3 = 2x + 18

Combining 8x - 3x gives 5x, and -12 + 15 gives +3.

4

Isolate x by subtracting 2x and 3 from both sides of the equation.

3x = 15 \implies x = 5

Subtracting 2x from both sides yields 3x + 3 = 18, then subtracting 3 yields 3x = 15, giving x = 5.

5

Substitute x = 5 into the expression 4x - 1 to find the final value.

4(5) - 1 = 20 - 1 = 19

The question asks for the evaluation of 4x - 1 rather than the value of x alone.

Key Concept

Solving linear equations in one variable involving fractional expressions and evaluating algebraic expressions.

Estimated Time:2m 0s

Question 5Question →

Two water pumps, Pump A and Pump B, were used to drain a reservoir containing 12,000 gallons of water. Pump A operates at a constant rate that is 50 gallons per hour greater than the rate of Pump B. Pump A worked alone for 4 hours, after which both pumps worked together for another 8 hours to completely empty the reservoir. What is the pumping rate of Pump B, in gallons per hour?

Show answer & explanation

Answer: 570

Answer

570

Let

r

represent the rate of Pump B in gallons per hour. Pump A's rate is

(r + 50)

gallons per hour. In the first 4 hours, Pump A drains

4(r + 50) = 4r + 200

gallons. In the next 8 hours, both pumps operate together at a combined rate of

(r + 50) + r = 2r + 50

gallons per hour, draining

8(2r + 50) = 16r + 400

gallons. Adding both quantities gives total volume drained:

(4r + 200) + (16r + 400) = 12,000

. Simplifying gives

20r + 600 = 12,000

, so

20r = 11,400

, which yields

r = 570

gallons per hour.

Step-by-Step Solution

1

Define variables for the rate of each pump

Rate of Pump B =

r

gal/hr; Rate of Pump A =

r + 50

gal/hr

Establishing a single unknown variable allows setting up a one-variable linear equation.

2

Write expressions for water drained during each time period

Period 1 (Pump A alone for 4 hrs):

4(r + 50) = 4r + 200

; Period 2 (Both pumps for 8 hrs):

8(2r + 50) = 16r + 400

Work done equals rate multiplied by time for each phase of operation.

3

Sum the work done in both periods to equal total volume and solve for

r

20r + 600 = 12,000 \Rightarrow 20r = 11,400 \Rightarrow r = 570

Solving the linear equation yields the exact rate of Pump B.

Key Concept

Linear Equations in One Variable

Estimated Time:2m 0s

Question 6Question →

Two commercial printing presses, Press A and Press B, operate at constant rates to print a total order of $N$ pages. Press A prints at a constant rate of $x$ pages per minute, while Press B prints at a constant rate that is $25$ pages per minute faster than Press A. Press A begins printing alone. After $20$ minutes, Press B is turned on, and both presses work simultaneously for an additional $30$ minutes. At that point, Press A stops, and Press B works alone for $10$ final minutes to complete the order. If Press B printed exactly $\frac{6}{11}$ of the total number of pages in the order, what is the value of $x$ ?

Show answer & explanation

Answer: 50

Answer

The value of

x

is 50.

To find

x

, calculate the total time each press operated. Press A ran for 20 minutes alone plus 30 minutes with Press B, giving 50 minutes total at

x

pages per minute (

50x

pages). Press B ran for 30 minutes with Press A plus 10 minutes alone, giving 40 minutes total at

(x + 25)

pages per minute (

40x + 1000

pages). The total order size is

N = 90x + 1000

. Setting Press B's output equal to

\frac{6}{11}N

yields

40x + 1000 = \frac{6}{11}(90x + 1000)

. Multiplying both sides by 11 gives

440x + 11000 = 540x + 6000

, which simplifies to

100x = 5000

, giving

x = 50

.

Step-by-Step Solution

1

Determine total operating times and express pages printed by each press in terms of

x

Press A printed

50x

pages; Press B printed

40(x + 25) = 40x + 1000

pages.

Press A operated for 20 minutes alone plus 30 minutes together (50 minutes total). Press B operated for 30 minutes together plus 10 minutes alone (40 minutes total).

2

Write the expression for total pages

N

N = 50x + (40x + 1000) = 90x + 1000

The total pages in the order is the sum of the pages printed by Press A and Press B.

3

Formulate the linear equation in one variable using the given ratio

40x + 1000 = \frac{6}{11}(90x + 1000)

Press B printed exactly

\frac{6}{11}

of the total pages

N

.

4

Solve the linear equation for

x

x = 50

Multiplying both sides by 11 clears the fraction to give

440x + 11000 = 540x + 6000

, which simplifies to

100x = 5000

.

Key Concept

Formulating and solving linear equations in one variable from multi-step rate and work scenarios.

Question 7Question →

At a charity fundraising event, standard tickets were sold for $\$45$ each and VIP tickets were sold for $\$80$ each. The number of standard tickets sold was $12$ more than twice the number of VIP tickets sold. If the total revenue generated from standard tickets exceeded the total revenue from VIP tickets by $\$1,830$ , how many VIP tickets were sold?

Show answer & explanation

Answer: 129

Answer

129 VIP tickets were sold.

Letting

v

represent the number of VIP tickets sold, the number of standard tickets sold is

2v + 12

. Expressing the revenue condition yields the linear equation

45(2v + 12) - 80v = 1830

. Distributing 45 gives

90v + 540 - 80v = 1830

. Combining like terms results in

10v + 540 = 1830

. Subtracting 540 from both sides gives

10v = 1290

, which yields

v = 129

.

Step-by-Step Solution

1

Define the unknown variable

Let

v

be the number of VIP tickets sold.

The question asks for the number of VIP tickets, making

v

a direct choice for the variable.

2

Translate the relationship between ticket quantities into an algebraic expression

Standard tickets sold

= 2v + 12

'12 more than twice the number of VIP tickets' translates directly to

2v + 12

.

3

Formulate total revenue expressions and construct the single-variable linear equation

45(2v + 12) - 80v = 1830

Total standard revenue minus total VIP revenue equals the given excess of

\$1,830

.

4

Distribute and combine like terms to solve for

v

90v + 540 - 80v = 1830 \implies 10v + 540 = 1830 \implies 10v = 1290 \implies v = 129

Applying standard algebraic operations isolates

v

on one side of the equation.

Key Concept

Formulating and solving a linear equation in one variable from a multi-step word problem context

Estimated Time:2m 0s

Question 8Question →

Two water reservoirs, Reservoir A and Reservoir B, contain a combined total of $360$ liters of water. After $30$ liters of water are transferred from Reservoir A to Reservoir B, the volume of water in Reservoir B is equal to $\frac{2}{3}$ of the volume of water remaining in Reservoir A. How many liters of water were originally in Reservoir A?

114

210

216

246

228

Show answer & explanation

Answer: $246$

Answer

246 liters were originally in Reservoir A.

Let

x

represent the original volume of Reservoir A. Since the combined volume is

360

liters, Reservoir B initially contains

360 - x

liters. After transferring

30

liters from A to B, Reservoir A contains

x - 30

liters and Reservoir B contains

(360 - x) + 30 = 390 - x

liters. Setting up the equation

390 - x = \frac{2}{3}(x - 30)

and multiplying by 3 gives

1170 - 3x = 2x - 60

. Combining like terms yields

5x = 1230

, so

x = 246

liters. Thus,

246

is the correct original volume.

Step-by-Step Solution

1

Define the variable for the unknown quantity.

Let

x

be the original volume of water in Reservoir A (in liters).

Choosing a single variable simplifies setting up a linear equation.

2

Express the original volume of Reservoir B in terms of

x

.

Original volume in Reservoir B is

360 - x

.

The total volume across both reservoirs is given as

360

liters.

3

Write expressions for the volumes in each reservoir after the transfer of 30 liters.

Volume in Reservoir A after transfer:

x - 30

.
Volume in Reservoir B after transfer:

(360 - x) + 30 = 390 - x

.

Transferring 30 liters removes 30 liters from A and adds 30 liters to B.

4

Set up the linear equation based on the given relationship.

390 - x = \frac{2}{3}(x - 30)

The problem specifies that the new volume in Reservoir B is

\frac{2}{3}

of the new volume in Reservoir A.

5

Solve the equation for

x

.

3(390 - x) = 2(x - 30) \implies 1170 - 3x = 2x - 60 \implies 1230 = 5x \implies x = 246

.

Multiply both sides by 3 to eliminate the fraction, then collect like terms.

Key Concept

Formulating and solving a linear equation in one variable from a real-world conservation/transfer scenario.

Question 9Question →

A manufacturing company's monthly total production cost

C

, in dollars, is related to the number of units produced,

x

, by the linear equation:

\frac{3(C - 1,200)}{5} = 2x + 180

If the total production cost in a given month was between

\$4,000

and

\$5,000

, inclusive, which of the following could be the number of units produced in that month? Select all such numbers.

Select all that apply

700

800

950

1,000

1,100

Show answer & explanation

Answer: 800; 950; 1,000

Answer

The numbers of units that could be produced are 800, 950, and 1,000.

Solving the linear equation for

x

gives

x = \frac{3C - 4,500}{10}

. Substituting the boundary cost values

C = 4,000

and

C = 5,000

yields

x = 750

and

x = 1,050

, respectively. Therefore, any valid number of units must lie in the range

[750, 1,050]

. The values 800, 950, and 1,000 fall strictly within this interval.

Step-by-Step Solution

1

Express x in terms of C by isolating the variable x in the linear equation.

3(C - 1,200) = 5(2x + 180) \implies 3C - 3,600 = 10x + 900 \implies 10x = 3C - 4,500 \implies x = \frac{3C - 4,500}{10}

Isolating

x

allows direct evaluation of the unit limits based on the cost boundary values.

2

Calculate the minimum value of x corresponding to the minimum cost C = 4,000.

x_{\min} = \frac{3(4,000) - 4,500}{10} = \frac{12,000 - 4,500}{10} = \frac{7,500}{10} = 750

Determines the lower bound of units produced.

3

Calculate the maximum value of x corresponding to the maximum cost C = 5,000.

x_{\max} = \frac{3(5,000) - 4,500}{10} = \frac{15,000 - 4,500}{10} = \frac{10,500}{10} = 1,050

Determines the upper bound of units produced.

4

Filter the given choices to find all values within the interval [750, 1,050].

The valid values are 800, 950, and 1,000.

Only values between 750 and 1,050 inclusive satisfy the cost constraint.

Key Concept

Linear Equations in One Variable

Question 10Question →

If $x$ satisfies the equation $\frac{x + 5}{3} - 2 = \frac{x - 1}{4}$ , what is the value of $x$ ?

-21

-7

1

7

17

Show answer & explanation

Answer: 1

Answer

1

Multiplying the equation

\frac{x + 5}{3} - 2 = \frac{x - 1}{4}

by the common denominator

12

eliminates fractions to produce

4(x + 5) - 24 = 3(x - 1)

. Expanding both sides yields

4x + 20 - 24 = 3x - 3

, which simplifies to

4x - 4 = 3x - 3

. Subtracting

3x

from both sides and adding

4

to both sides gives

x = 1

.

Step-by-Step Solution

1

Multiply every term on both sides of the equation by the least common denominator of 3 and 4, which is 12.

12 \cdot \left(\frac{x + 5}{3}\right) - 12 \cdot 2 = 12 \cdot \left(\frac{x - 1}{4}\right)

, which simplifies to

4(x + 5) - 24 = 3(x - 1)

.

Clearing fractional denominators simplifies the equation into integer-coefficient linear form.

2

Expand both sides by distributing the numeric multipliers.

4x + 20 - 24 = 3x - 3

, which combines like terms to

4x - 4 = 3x - 3

.

Distributing coefficients removes grouping symbols so variable and constant terms can be combined.

3

Isolate the variable term

x

on one side.

Subtract

3x

from both sides to get

x - 4 = -3

, then add

4

to both sides to get

x = 1

.

Standard algebraic reduction requires grouping all terms containing the unknown variable on one side and numerical constants on the other.

Key Concept

Solving single-variable linear equations containing fractional terms by clearing denominators

Estimated Time:1m 0s

Question 11Question →

If $x$ satisfies the linear equation $\frac{x + 2}{2} + \frac{x - 1}{3} = 4$ , what is the value of $3x - 2$ ?

10

4

-2

26.5

7.6

Show answer & explanation

Answer: $10$

Answer

The value of

3x - 2

is

10

.

To solve

\frac{x + 2}{2} + \frac{x - 1}{3} = 4

, multiply the entire equation by the common denominator 6 to clear fractions, yielding

3(x + 2) + 2(x - 1) = 24

. Distributing gives

3x + 6 + 2x - 2 = 24

, which simplifies to

5x + 4 = 24 \implies 5x = 20 \implies x = 4

. Substituting

x = 4

into

3x - 2

gives

3(4) - 2 = 10

.

Step-by-Step Solution

1

Find a common denominator to clear the fractions from the equation.

The least common multiple of 2 and 3 is 6. Multiplying both sides of the equation by 6 gives

6 \cdot \left(\frac{x + 2}{2}\right) + 6 \cdot \left(\frac{x - 1}{3}\right) = 6 \cdot 4

, which simplifies to

3(x + 2) + 2(x - 1) = 24

.

Clearing denominators simplifies multi-step fractional equations into standard linear form.

2

Expand terms and solve for

x

.

3x + 6 + 2x - 2 = 24 \implies 5x + 4 = 24 \implies 5x = 20 \implies x = 4

.

Combining like terms isolates the variable

x

.

3

Substitute

x = 4

into the target expression

3x - 2

.

3(4) - 2 = 12 - 2 = 10

.

The question asks for the value of the algebraic expression

3x - 2

, not

x

itself.

Key Concept

Solving linear equations with fractional coefficients by clearing denominators and evaluating target algebraic expressions.

Estimated Time:1m 15s

Question 12Question →

A shipping company calculates the total cost $C$ , in dollars, to deliver a package of weight $w$ pounds using the linear relationship $C = kw + b$ , where $k$ and $b$ are positive constants. The delivery cost for a $4$ -pound package is $\$19$ , and the delivery cost for a $9$ -pound package is $\$39$ . If a customer pays a total delivery cost of $\$71$ for a single package, what is the weight of the package, in pounds?

13

15

17

18

21

Show answer & explanation

Answer: $17$

Answer

The weight of the package is

17

pounds.

Using the two points

(4, 19)

and

(9, 39)

, the rate of change

k

is calculated as

\frac{39 - 19}{9 - 4} = \frac{20}{5} = 4

dollars per pound. Substituting

k = 4

into

4(4) + b = 19

yields

b = 3

. Setting the linear equation

4w + 3 = 71

and solving for

w

gives

4w = 68

, so

w = 17

pounds.

Step-by-Step Solution

1

Set up a system of linear equations using the given data points

(4, 19)

and

(9, 39)

.

4k + b = 19

and

9k + b = 39

.

The cost model follows

C = kw + b

for weight

w

.

2

Subtract the first equation from the second equation to solve for

k

.

(9k + b) - (4k + b) = 39 - 19 \implies 5k = 20 \implies k = 4

.

Subtracting eliminates the constant

b

to determine the unit rate per pound.

3

Substitute

k = 4

into

4k + b = 19

to solve for

b

.

4(4) + b = 19 \implies 16 + b = 19 \implies b = 3

.

Finding

b

establishes the full linear equation model:

C = 4w + 3

.

4

Substitute

C = 71

into the linear equation

4w + 3 = 71

and solve for

w

.

4w = 71 - 3 \implies 4w = 68 \implies w = 17

.

Solving for

w

yields the required weight corresponding to a

\$71

delivery cost.

Key Concept

Linear Modeling and Single-Variable Linear Equations

Estimated Time:1m 30s

Question 13Question →

If $\frac{x}{2} + \frac{x}{3} = 10$ , what is the value of $x$ ?

25

12

2

6

15

Show answer & explanation

Answer: 12

Answer

The value of

x

is 12.

The option showing 12 is correct because converting the fractions

\frac{x}{2}

and

\frac{x}{3}

to have a common denominator of 6 produces

\frac{3x + 2x}{6} = 10

, or

\frac{5x}{6} = 10

. Multiplying both sides by 6 gives

5x = 60

, and dividing by 5 yields

x = 12

.

Step-by-Step Solution

1

Find a common denominator for the fractional terms on the left side of the equation.

The least common denominator of 2 and 3 is 6. Rewriting the fractions gives

\frac{3x}{6} + \frac{2x}{6} = 10

.

Fractions must have identical denominators before their numerators can be added.

2

Combine the fractions on the left side.

\frac{5x}{6} = 10

Adding the numerators

3x + 2x

yields

5x

over the shared denominator 6.

3

Clear the denominator by multiplying both sides of the equation by 6.

5x = 60

Multiplying both sides by 6 eliminates the fraction while maintaining equality.

4

Isolate

x

by dividing both sides by 5.

x = 12

Dividing 60 by 5 yields the solution for

x

.

Key Concept

Solving linear equations involving fractions by clearing denominators using the least common multiple.

Estimated Time:45s

Question 14Question →

A company allocated a total budget of $B$ dollars for a project. In the first phase of the project, $\frac{2}{5}$ of the total budget plus $\$3,000$ was spent. In the second phase, $\frac{1}{3}$ of the remaining budget after the first phase was spent. If the unspent amount after both phases is $\$14,000$ , what was the total initial budget $B$ ?

\$27,200

\$30,000

\$40,000

\$42,500

\$75,000

Show answer & explanation

Answer: $\$40,000$

Answer

\$40,000

The correct answer of

\$40,000

is obtained by systematically tracking the remaining budget after each phase. After Phase 1, the remaining amount is

B - (\frac{2}{5}B + 3,000) = \frac{3}{5}B - 3,000

. Spending

\frac{1}{3}

of this balance in Phase 2 leaves

\frac{2}{3}

of it unspent:

\frac{2}{3}(\frac{3}{5}B - 3,000) = \frac{2}{5}B - 2,000

. Setting this equal to the final unspent amount of

\$14,000

gives

\frac{2}{5}B = 16,000

, which solves to

B = 40,000

.

Step-by-Step Solution

1

Express the remaining budget after the first phase in terms of

B

.

Amount spent in Phase 1 =

\frac{2}{5}B + 3,000

. Remaining after Phase 1 =

B - \left(\frac{2}{5}B + 3,000\right) = \frac{3}{5}B - 3,000

.

Subtracting the first phase expenses from the initial total budget

B

determines the balance available for the second phase.

2

Express the unspent budget after the second phase.

Since

\frac{1}{3}

of the remaining budget was spent in Phase 2,

1 - \frac{1}{3} = \frac{2}{3}

of that remaining budget is left. Remaining after Phase 2 =

\frac{2}{3}\left(\frac{3}{5}B - 3,000\right)

.

Taking

\frac{2}{3}

of the Phase 1 remainder directly gives the final unspent amount.

3

Expand and simplify the algebraic equation setting the unspent amount equal to

\$14,000

.

\frac{2}{3}\left(\frac{3}{5}B - 3,000\right) = 14,000 \implies \frac{2}{5}B - 2,000 = 14,000.

Distributing

\frac{2}{3}

across both terms inside the parentheses clears the fraction product.

4

Solve the linear equation for

B

.

\frac{2}{5}B = 16,000 \implies B = 16,000 \times \frac{5}{2} = 40,000.

Adding

2,000

to both sides and multiplying by the reciprocal

\frac{5}{2}

yields the total budget

B

.

Key Concept

Formulating and solving multi-step linear equations in one variable with fractional quantities and consecutive remaining balances.

Estimated Time:2m 0s

Question 15Question →

A company's annual budget of $\$84,000$ is split among three departments: Research, Marketing, and Operations. The Marketing department receives $\frac{2}{3}$ as much funding as the Research department. The Operations department receives $\$6,000$ more than half of the combined funding of the Research and Marketing departments. What is the amount, in dollars, allocated to the Research department?

24,000

28,800

31,200

33,600

36,000

Show answer & explanation

Answer: $31,200$

Answer

31,200

Defining Research funding as

x

, Marketing funding becomes

\frac{2}{3}x

, and Operations funding becomes

\frac{1}{2}(x + \frac{2}{3}x) + 6000 = \frac{5}{6}x + 6000

. Summing all three department allocations yields

x + \frac{2}{3}x + \frac{5}{6}x + 6000 = 84,000

. Combining the variable terms gives

\frac{5}{2}x + 6000 = 84,000

, which simplifies to

\frac{5}{2}x = 78,000

and yields

x = 31,200

.

Step-by-Step Solution

1

Define the unknown variable for the target quantity.

Let

x

represent the dollar amount allocated to the Research department.

The problem asks specifically for the Research department allocation.

2

Express the allocations of Marketing and Operations in terms of

x

.

Marketing

= \frac{2}{3}x

. Combined Research and Marketing

= x + \frac{2}{3}x = \frac{5}{3}x

. Operations

= \frac{1}{2}\left(\frac{5}{3}x\right) + 6,000 = \frac{5}{6}x + 6,000

.

Translating word problem relationships into algebraic expressions.

3

Set up the single-variable linear equation for the total budget.

x + \frac{2}{3}x + \left(\frac{5}{6}x + 6,000\right) = 84,000

The sum of allocations across all three departments must equal the total budget of

\$84,000

.

4

Combine like terms using a common denominator.

\frac{6}{6}x + \frac{4}{6}x + \frac{5}{6}x + 6,000 = 84,000 \implies \frac{15}{6}x + 6,000 = 84,000 \implies \frac{5}{2}x + 6,000 = 84,000

Simplifying fractional coefficients by finding the common denominator 6.

5

Isolate

x

to solve the linear equation.

\frac{5}{2}x = 78,000 \implies 5x = 156,000 \implies x = 31,200

Subtracting

6,000

from both sides and multiplying by

\frac{2}{5}

.

Key Concept

Linear Equations in One Variable

Question 16Question →

For a constant

a

, consider the following linear equation in one variable

x

:

\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + 5}{12}

Which of the following statements are true? Select all such statements.

Select all that apply

If

a = 1

, the equation has no solution for

x

.

If

a = 5

, then

x = 3

.

If

a = -1

, then

x = 0

.

If

x = 4

, then

a = 5

.

If

a = 0

, the equation has a positive solution for

x

.

Show answer & explanation

Answer: If $a = 1$ , the equation has no solution for $x$ .; If $a = -1$ , then $x = 0$ .; If $x = 4$ , then $a = 5$ .

Answer

The correct statements are the ones asserting that

a = 1

leaves the equation with no solution,

a = -1

leads to

x = 0

, and

x = 4

requires

a = 5

.

Multiplying the equation by 12 and simplifying yields

3(a - 1)x = 8(a + 1)

. When

a = 1

, the left side becomes 0 while the right side becomes 16, creating an inconsistent equation

0 = 16

with no solution. When

a = -1

, the equation becomes

-6x = 0

, which gives

x = 0

. When

x = 4

, substituting into the simplified equation gives

12(a - 1) = 8(a + 1)

, which simplifies to

4a = 20 \implies a = 5

.

Step-by-Step Solution

1

Clear denominators by multiplying every term of the equation by the least common multiple, 12.

4a(x - 2) - 3(2x + 1) = (a - 3)x + 5

Eliminating fractional coefficients simplifies the distribution and collection of variable terms.

2

Expand products on both sides of the equation.

4ax - 8a - 6x - 3 = ax - 3x + 5

Applying the distributive property isolates individual algebraic terms.

3

Rearrange terms to collect all

x

-terms on the left side and constant terms on the right side.

(4a - 6 - a + 3)x = 8a + 8 \implies 3(a - 1)x = 8(a + 1)

Factoring out

x

provides the canonical linear form

A x = B

.

4

Evaluate the given conditions for

a

and

x

against the canonical form

3(a - 1)x = 8(a + 1)

.

For

a = 1

:

0 = 16

(no solution). For

a = 5

:

12x = 48 \implies x = 4

. For

a = -1

:

-6x = 0 \implies x = 0

. For

x = 4

:

12(a - 1) = 8(a + 1) \implies a = 5

. For

a = 0

:

-3x = 8 \implies x = -\frac{8}{3}

(negative).

Direct substitution verifies which algebraic relationships hold true.

Key Concept

Linear equations in one variable containing symbolic parameters can be analyzed for existence of solutions, zero-roots, and explicit values by reducing to the form

A x = B

.

Estimated Time:2m 30s

Question 17Question →

A container originally holds a liquid mixture consisting of substance A and substance B, where substance A constitutes $\frac{2}{5}$ of the total volume. After $15$ liters of substance B are added to the container and $3$ liters of substance A evaporate, the volume of substance A in the container becomes $\frac{1}{4}$ of the new total liquid volume. What was the original total volume, in liters, of the liquid mixture in the container?

32

40

45

50

52

Show answer & explanation

Answer: 40

Answer

40 liters

Letting

V

represent the original volume of the liquid mixture in liters, the initial amount of substance A is

\frac{2}{5}V

. After adding 15 liters of substance B and losing 3 liters of substance A to evaporation, the updated volume of substance A is

\frac{2}{5}V - 3

, and the updated total volume is

V + 15 - 3 = V + 12

. Setting up the relationship

\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)

and expanding the right side gives

\frac{2}{5}V - 3 = \frac{1}{4}V + 3

. Subtracting

\frac{1}{4}V

from both sides yields

\frac{3}{20}V = 6

, which solves to

V = 40

liters.

Step-by-Step Solution

1

Define the unknown variable and express initial quantities algebraically.

Let

V

be the original total volume of the liquid mixture in liters. The original volume of substance A is

\frac{2}{5}V

.

Establishing a variable for the initial total volume allows all changes to be modeled in terms of one variable.

2

Express the modified quantities after additions and evaporation.

New volume of substance A

= \frac{2}{5}V - 3

. New total volume

= V + 15 - 3 = V + 12

.

Adding 15 liters of substance B increases the total volume by 15, and losing 3 liters of substance A decreases both substance A and the total volume by 3.

3

Set up the linear equation based on the given ratio condition.

\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)

Substance A forms one-fourth of the updated total liquid volume.

4

Expand and solve the linear equation for

V

.

\frac{2}{5}V - 3 = \frac{1}{4}V + 3 \implies \frac{2}{5}V - \frac{1}{4}V = 6 \implies \frac{8 - 5}{20}V = 6 \implies \frac{3}{20}V = 6 \implies V = 40.

Clearing terms and subtracting

\frac{1}{4}V

from

\frac{2}{5}V

gives

\frac{3}{20}V = 6

, which yields

V = 40

.

Key Concept

Linear Equations in One Variable

Question 18Question →

If

x

satisfies the linear equation

\frac{3(x - 2)}{4} - \frac{x - 5}{3} = \frac{2x + 1}{6} + 2

what is the value of

2x - 5

?

43

24

123

-1

51

Show answer & explanation

Answer: 43

Answer

43

The correct answer is obtained by clearing denominators with the least common denominator 12, yielding

9(x - 2) - 4(x - 5) = 2(2x + 1) + 24

. Expanding and simplifying gives

5x + 2 = 4x + 26

, which isolates

x = 24

. Substituting

x = 24

into

2x - 5

results in

2(24) - 5 = 43

.

Step-by-Step Solution

1

Clear the denominators by multiplying every term in the equation by the least common multiple of 4, 3, and 6, which is 12.

12 \cdot \frac{3(x - 2)}{4} - 12 \cdot \frac{x - 5}{3} = 12 \cdot \frac{2x + 1}{6} + 12 \cdot 2 \implies 9(x - 2) - 4(x - 5) = 2(2x + 1) + 24

Eliminating fractions simplifies the linear equation into standard polynomial form.

2

Distribute the constants through the parentheses, taking careful note of negative signs.

9x - 18 - 4x + 20 = 4x + 2 + 24

Parentheses must be removed to collect like variable and constant terms.

3

Combine like terms on both sides of the equation.

5x + 2 = 4x + 26

Consolidating terms on each side allows for isolating the variable.

4

Isolate

x

on one side of the equation.

5x - 4x = 26 - 2 \implies x = 24

Subtracting

4x

and 2 from both sides solves directly for

x

.

5

Evaluate the target expression

2x - 5

using

x = 24

.

2(24) - 5 = 48 - 5 = 43

The question requests the value of

2x - 5

, not

x

alone.

Key Concept

Solving multi-step linear equations in one variable with fractional terms and evaluating algebraic expressions.

Estimated Time:2m 0s

Question 19Question →

If

x

satisfies the linear equation

\frac{3x - 1}{4} - \frac{2x + 3}{5} = \frac{x - 3}{4} + \frac{2}{5}

what is the value of

4x - 3

?

5

13

17

57

77

Show answer & explanation

Answer: 17

Answer

17

Solving the equation by grouping terms with like denominators gives

\frac{2x + 2}{4} = \frac{2x + 5}{5}

. Simplifying the left side to

\frac{x + 1}{2}

and cross-multiplying yields

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

, which simplifies to

5x + 5 = 4x + 10

, giving

x = 5

. Evaluating

4x - 3

at

x = 5

produces

4(5) - 3 = 17

, making this answer correct.

Step-by-Step Solution

1

Group like fractional terms with denominator 4 on one side and denominator 5 on the other side of the equation.

\frac{3x - 1}{4} - \frac{x - 3}{4} = \frac{2x + 3}{5} + \frac{2}{5}

Grouping terms with common denominators simplifies algebraic combination.

2

Combine the numerators over their common denominators, carefully distributing signs.

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

Subtracting

(x - 3)

requires distributing the negative sign to yield

-x + 3

.

3

Simplify the left side fraction and cross-multiply to eliminate denominators.

\frac{x + 1}{2} = \frac{2x + 5}{5} \implies 5(x + 1) = 2(2x + 5)

Simplifying

\frac{2x+2}{4}

to

\frac{x+1}{2}

reduces computation before cross-multiplication.

4

Expand both sides and solve for

x

.

5x + 5 = 4x + 10 \implies 5x - 4x = 10 - 5 \implies x = 5

Subtracting

4x

and

5

from both sides isolates

x

.

5

Substitute

x = 5

into the required expression

4x - 3

.

4(5) - 3 = 20 - 3 = 17

The question asks for the value of

4x - 3

, not

x

alone.

Key Concept

Solving linear equations in one variable involving fractional terms and evaluating targeted algebraic expressions.

Estimated Time:2m 0s

Question 20Question →

A corporate enterprise allocates a total annual budget of $\$52,000$ across three departments: Marketing, Operations, and Development. The amount allocated to Operations is $\$5,000$ less than twice the amount allocated to Marketing. The amount allocated to Development is $\$1,000$ more than half of the combined allocations of Marketing and Operations. What is the budget allocation for Development?

\$13,000

\$17,000

\$18,000

\$21,000

\$24,000

Show answer & explanation

Answer: $\$18,000$

Answer

The budget allocation for Development is

\$18,000

.

By defining the Marketing budget as

x

, Operations becomes

2x - 5,000

and Development becomes

\frac{1}{2}(3x - 5,000) + 1,000 = 1.5x - 1,500

. Adding these three expressions equals the total budget of

\$52,000

, giving

4.5x - 6,500 = 52,000

. Solving for

x

yields

x = 13,000

. Substituting

x = 13,000

into the expression for Development gives

1.5(13,000) - 1,500 = \$18,000

.

Step-by-Step Solution

1

Define the variable and express each department's budget in terms of that variable.

Let

x

be the Marketing budget in dollars. Then Operations

= 2x - 5,000

. The combined Marketing and Operations budget

= x + (2x - 5,000) = 3x - 5,000

. Therefore, Development

= \frac{1}{2}(3x - 5,000) + 1,000 = 1.5x - 1,500

.

Establishing a single linear variable allows all three department budgets to be combined into one linear equation.

2

Set up the total budget equation and solve for

x

.

x + (2x - 5,000) + (1.5x - 1,500) = 52,000 \implies 4.5x - 6,500 = 52,000 \implies 4.5x = 58,500 \implies x = 13,000

.

Summing the allocations for Marketing, Operations, and Development yields the total corporate budget of

\$52,000

.

3

Calculate the specific budget allocation requested for Development.

Development allocation

= 1.5(13,000) - 1,500 = 19,500 - 1,500 = 18,000

.

Substituting

x = 13,000

back into the algebraic expression for Development gives the final required dollar amount.

Key Concept

Formulating and solving multi-step linear equations in one variable from word problems.

Estimated Time:2m 30s

Linear Equations in One Variable

Answer

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Key Concept

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