Linear Equations in One Variable Practice Questions — SAT — Page 4

Question 61Question →

Oven A preheats at a constant rate of $15$ degrees Celsius per minute, starting from an initial temperature of $25$ degrees Celsius. Oven B starts preheating $5$ minutes after Oven A begins, starting from an initial temperature of $20$ degrees Celsius and preheating at a constant rate of $20$ degrees Celsius per minute. If both ovens continue to preheat, how many minutes after Oven A begins preheating will both ovens reach the same temperature?

Show answer & explanation

Answer: 21

Answer

The ovens will reach the same temperature

21

minutes after Oven A begins preheating.

To find the number of minutes after Oven A begins preheating when both ovens reach the same temperature, we can write an equation in terms of

t

, the time in minutes since Oven A started. Oven A starts at

25

degrees Celsius and increases by

15

degrees per minute, so its temperature is

25 + 15t

. Oven B starts

5

minutes later, meaning it preheats for

t - 5

minutes. Starting from

20

degrees Celsius and preheating at

20

degrees per minute, Oven B's temperature is

20 + 20(t - 5)

. Setting these two expressions equal gives the equation

25 + 15t = 20 + 20(t - 5)

. Distributing

20

yields

25 + 15t = 20t - 80

. Isolating

t

gives

5t = 105

, which results in

t = 21

.

Step-by-Step Solution

1

Set up expressions representing the temperature of each oven

t

minutes after Oven A begins preheating.

Oven A:

25 + 15t

; Oven B:

20 + 20(t - 5)

Since Oven B starts

5

minutes after Oven A, it has been preheating for

t - 5

minutes.

2

Set the temperature expressions equal to find the time at which they reach the same temperature.

25 + 15t = 20 + 20(t - 5)

We want to find the value of

t

where the temperatures of the two ovens are equal.

3

Distribute and simplify the equation.

25 + 15t = 20t - 80

Distributing the

20

across

(t - 5)

yields

20t - 100

. Combining the constant terms gives

20 - 100 = -80

.

4

Solve for

t

by isolating the variable term.

5t = 105

, which gives

t = 21

Subtract

15t

and add

80

to both sides, then divide by

5

.

Key Concept

Solving linear equations in one variable that model real-world situations with a time delay.

Question 62Question →

In the equation below, $a$ and $b$ are constants.

a(2x - 9) - 3(x - b) = 7x + 12

If the equation has infinitely many solutions, what is the value of $b$ ?

-19

10

11

19

Show answer & explanation

Answer: 19

Answer

19

To find the value of

b

for which the equation has infinitely many solutions, we expand the left side to get

2ax - 9a - 3x + 3b = 7x + 12

. Grouping the terms by variable yields

(2a - 3)x + (3b - 9a) = 7x + 12

. For the equation to have infinitely many solutions, the coefficient of

x

and the constant term on both sides must be equal. Setting the

x

-coefficients equal gives

2a - 3 = 7

, which simplifies to

2a = 10

and

a = 5

. Setting the constant terms equal gives

3b - 9a = 12

. Substituting

a = 5

into this equation yields

3b - 9(5) = 12

, which simplifies to

3b - 45 = 12

. Adding

45

to both sides gives

3b = 57

, which results in

b = 19

.

Step-by-Step Solution

1

Expand and simplify the left side of the equation.

(2a - 3)x + (3b - 9a) = 7x + 12

To compare coefficients with the right side of the equation, we need to group the terms on the left side.

2

Set the coefficients of

x

on both sides equal to each other to solve for

a

.

2a - 3 = 7 \implies 2a = 10 \implies a = 5

For an equation to have infinitely many solutions, the coefficients of the variable on both sides must be equal.

3

Set the constant terms on both sides equal to each other, substitute the value of

a

, and solve for

b

.

3b - 9a = 12 \implies 3b - 9(5) = 12 \implies 3b - 45 = 12 \implies 3b = 57 \implies b = 19

For an equation to have infinitely many solutions, the constant terms on both sides must also be equal.

Key Concept

Linear equations with infinitely many solutions require the coefficients of the variable to be equal and the constant terms to be equal on both sides of the equation.

Question 63Question →

In the equation below, $k$ is a constant.

15x - 2(3x - k) = 9x + 10

If the equation has infinitely many solutions, what is the value of $k$ ?

-5

20

5

8

Show answer & explanation

Answer: 5

Answer

The correct value of

k

is

5

, which makes the equation have infinitely many solutions.

The correct answer is

5

because distributing

-2

through the parentheses on the left side of the equation yields

15x - 6x + 2k = 9x + 10

. Combining the

x

terms gives

9x + 2k = 9x + 10

. For a linear equation in one variable to have infinitely many solutions, both sides must be identical. Since the coefficients of

x

are already equal (

9 = 9

), we set the constants equal to each other (

2k = 10

), which simplifies to

k = 5

.

Step-by-Step Solution

1

Distribute the constant

-2

to both terms inside the parentheses on the left side of the equation.

15x - 6x + 2k = 9x + 10

Applying the distributive property correctly simplifies the terms inside the parentheses.

2

Combine the like terms of

x

on the left side of the equation.

9x + 2k = 9x + 10

Subtracting

6x

from

15x

simplifies the left side of the equation to have a single

x

term.

3

Set the constant terms on both sides equal to each other to find the condition for infinitely many solutions.

2k = 10

For a linear equation in one variable to have infinitely many solutions, the coefficients of the variable on both sides must be equal, and the constant terms on both sides must also be equal.

4

Solve for

k

by dividing both sides of the equation by

2

.

k = 5

Isolating the variable

k

yields the final value.

Key Concept

For a linear equation in one variable to have infinitely many solutions, the equation must simplify to an identity where the variable coefficients are equal and the constant terms are equal on both sides of the equation.

Estimated Time:1m 15s

Question 64Question →

A customer opens a savings account with an initial deposit of $\$180$ and deposits $\$30$ at the end of each week. Two weeks later, a second customer opens a savings account with an initial deposit of $\$120$ and deposits $\$50$ at the end of each week. If neither customer makes any other deposits or withdrawals, after how many weeks from the time the first customer opened their account will both accounts have the same balance?

Show answer & explanation

Answer: 8

Answer

Both accounts will have the same balance after 8 weeks.

The correct answer is 8. Letting

w

represent the number of weeks since the first customer opened their account, the balance of the first account can be modeled by the expression

180 + 30w

. Since the second customer opens their account two weeks later, the number of weeks they have been depositing money is

w - 2

. Therefore, the balance of the second account can be modeled by the expression

120 + 50(w - 2)

. Setting the two expressions equal to find when the balances are the same gives the equation

180 + 30w = 120 + 50(w - 2)

. Distributing the

50

yields

180 + 30w = 120 + 50w - 100

. Simplifying the right side gives

180 + 30w = 50w + 20

. Subtracting

30w

and

20

from both sides of the equation yields

160 = 20w

. Dividing both sides by

20

gives

w = 8

.

Step-by-Step Solution

1

Set up expressions for the balance of each account after

w

weeks.

First account balance:

180 + 30w

; Second account balance:

120 + 50(w - 2)

The first customer has been depositing for

w

weeks. The second customer started 2 weeks later, so they have been depositing for

w - 2

weeks.

2

Set the two expressions equal to each other to find when the balances are equal.

180 + 30w = 120 + 50(w - 2)

We want to find the number of weeks

w

when the two account balances are equal.

3

Solve the linear equation for

w

.

w = 8

Distribute the 50 to get

180 + 30w = 120 + 50w - 100

. Combine constants to get

180 + 30w = 50w + 20

. Subtract

30w

and

20

from both sides to get

160 = 20w

. Divide by

20

to get

w = 8

.

Key Concept

Formulating and solving linear equations in one variable from real-world contexts

Question 65Question →

If $\frac{3}{5}(5x - 10) - \frac{1}{2}(4x - 6) = 12$ , what is the value of $2x - 5$ ?

13

15

25

37

Show answer & explanation

Answer: 25

Answer

The correct answer is 25.

The correct answer is 25. Distributing the fractions yields

3x - 6 - 2x + 3 = 12

, which simplifies to

x - 3 = 12

, so

x = 15

. Substituting this into

2x - 5

gives

2(15) - 5 = 25

.

Step-by-Step Solution

1

Distribute the coefficients to the terms inside the parentheses.

(3x - 6) - (2x - 3) = 12

To simplify the equation by removing the parentheses.

2

Combine like terms on the left side of the equation.

x - 3 = 12

To group the variable terms and constant terms together.

3

Isolate the variable

x

by adding 3 to both sides of the equation.

x = 15

To find the value of

x

.

4

Substitute

x = 15

into the expression

2x - 5

.

2(15) - 5 = 25

To find the final value requested by the question.

Key Concept

Solving linear equations in one variable with grouping symbols and fractions, and evaluating an algebraic expression.

Estimated Time:1m 30s

Question 66Question →

A subscription-based meal kit service charges a monthly membership fee of $\$35$ plus $\$8.50$ per meal. A non-member can purchase the same meals for $\$12.00$ each, but must pay a flat monthly delivery fee of $\$7$ . How many meals must be purchased in a month for the total monthly cost for a member to be equal to the total monthly cost for a non-member?

Show answer & explanation

Answer: 8

Answer

The total monthly cost is equal for a member and a non-member when 8 meals are purchased in a month.

The correct answer of 8 is found by setting the member cost expression,

35 + 8.5m

, equal to the non-member cost expression,

12m + 7

, and solving for the number of meals,

m

.

Step-by-Step Solution

1

Represent the total monthly cost for a member and a non-member using equations where

m

is the number of meals purchased.

Member cost is represented by

35 + 8.50m

, and non-member cost is represented by

12.00m + 7

.

To set up an algebraic representation of the cost structures.

2

Equate the two cost expressions to find the number of meals where the costs are equal.

35 + 8.50m = 12.00m + 7

To find the value of

m

that makes both costs equivalent.

3

Isolate the variable

m

on one side of the equation.

3.50m = 28

Subtracting

8.50m

and

7

from both sides simplifies the equation by keeping variable terms on one side and constants on the other.

4

Divide both sides of the equation by

3.50

to solve for

m

.

m = 8

Dividing

28

by

3.50

yields the final number of meals.

Key Concept

Solving linear equations in one variable with variable terms on both sides of the equation

Estimated Time:1m 30s

Question 67Question →

A laboratory technician is cooling a liquid sample. The initial temperature of the sample is $80^\circ\text{C}$ . The technician uses a cooling program that decreases the temperature at a constant rate of $1.5^\circ\text{C}$ per minute. After $t$ minutes, the technician increases the cooling rate by $0.75^\circ\text{C}$ per minute and runs the cooling program for another $12$ minutes. If the final temperature of the sample is $44^\circ\text{C}$ , what is the value of $t$ ?

8

6

18

42

Show answer & explanation

Answer: 6

Answer

6

The correct answer is

6

. The initial temperature of the sample is

80^\circ\text{C}

. For the first

t

minutes, the temperature decreases by

1.5t^\circ\text{C}

. For the next

12

minutes, the rate is

1.5 + 0.75 = 2.25^\circ\text{C}

per minute, resulting in a temperature decrease of

2.25 \times 12 = 27^\circ\text{C}

. The final temperature equation is

80 - 1.5t - 27 = 44

, which simplifies to

53 - 1.5t = 44

. Solving for

t

gives

1.5t = 9

, or

t = 6

.

Step-by-Step Solution

1

Determine the cooling rate for both phases of the program.

The initial rate is

1.5^\circ\text{C}

per minute. The increased rate is

1.5 + 0.75 = 2.25^\circ\text{C}

per minute.

The rate in the second phase is increased by

0.75^\circ\text{C}

per minute from the initial rate.

2

Set up the linear equation representing the temperature change over time.

80 - 1.5t - 2.25(12) = 44

The final temperature is the initial temperature minus the temperature drops during each phase.

3

Solve the equation to isolate the variable

t

.

80 - 1.5t - 27 = 44 \implies 53 - 1.5t = 44 \implies -1.5t = -9 \implies t = 6

Perform basic algebraic operations to find the value of

t

.

Key Concept

Setting up and solving a linear equation in one variable from a real-world scenario.

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