College Algebra Quiz: Solve Systems Of Linear Equations Substitution

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Solve Systems Of Linear Equations SubstitutionQuestion 1 of 9

Consider the system: $\frac{x}{2} + \frac{y}{3} = 5$ and $\frac{x}{4} - \frac{y}{6} = 1$ . When using substitution, if you solve the first equation for $x$ in terms of $y$ , what expression do you substitute into the second equation?

x = 30 - 2y

, obtained by multiplying the first equation by 6 and solving for

x

x = 5 - \frac{y}{3}

, obtained by multiplying the first equation by 2 and solving for

x

x = 15 - y

, obtained by multiplying the first equation by 3 and solving for

x

x = 10 - \frac{2y}{3}

, obtained by multiplying the first equation by 6 and solving for

x

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College Algebra Quiz

College Algebra Quiz: Solve Systems Of Linear Equations Substitution

Practice Solve Systems Of Linear Equations Substitution in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solve Systems Of Linear Equations Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$x = 30 - 2y$ , obtained by multiplying the first equation by 6 and solving for $x$
$x = 5 - \frac{y}{3}$ , obtained by multiplying the first equation by 2 and solving for $x$
$x = 15 - y$ , obtained by multiplying the first equation by 3 and solving for $x$
$x = 10 - \frac{2y}{3}$ , obtained by multiplying the first equation by 6 and solving for $x$ (correct answer)

Explanation: When solving systems of equations with fractions using substitution, you need to isolate one variable in terms of the other, then substitute that expression into the second equation. The key is handling the fractions correctly through algebraic manipulation. To solve the first equation

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

for

x

, you should first clear the fractions by multiplying the entire equation by the least common multiple of the denominators (2 and 3), which is 6. This gives you:

6 \cdot \frac{x}{2} + 6 \cdot \frac{y}{3} = 6 \cdot 5

, which simplifies to

3x + 2y = 30

. Now isolating

x

3x = 30 - 2y

, so

x = 10 - \frac{2y}{3}

. This matches answer choice D. Answer choice A gives

x = 30 - 2y

, which incorrectly omits dividing by 3 when solving

3x = 30 - 2y

for

x

. Answer choice B multiplies by 2 instead of 6, leaving you with

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

, leading to the incorrect expression

x = 5 - \frac{y}{3}

(this also contains an error in the fraction manipulation). Answer choice C multiplies by 3 instead of 6, giving

\frac{3x}{2} + y = 15

, which leads to

x = 15 - y

after incorrect simplification. When working with fractional coefficients in systems, always multiply by the LCM of all denominators first to eliminate fractions completely. This prevents the algebraic errors that create most wrong answer choices in substitution problems.

Question 2

Consider the system of equations: $3x - 2y = 7$ and $x + 4y = -1$ . When solving by substitution, if you solve the second equation for $x$ and substitute into the first equation, what equation in terms of $y$ do you obtain?

$3(-1 - 4y) - 2y = 7$ , which simplifies to $-14y = 10$ (correct answer)
$3(-1 + 4y) - 2y = 7$ , which simplifies to $10y = 10$
$3(1 - 4y) - 2y = 7$ , which simplifies to $-14y = 4$
$3(1 + 4y) - 2y = 7$ , which simplifies to $10y = 4$

Explanation: From

x + 4y = -1

, solving for

x

gives

x = -1 - 4y

. Substituting into

3x - 2y = 7

3(-1 - 4y) - 2y = 7

, which becomes

-3 - 12y - 2y = 7

, so

-14y = 10

. Choice B incorrectly writes

x = -1 + 4y

(sign error). Choice C incorrectly writes

x = 1 - 4y

(constant sign error). Choice D combines both sign errors.

Question 3

Two numbers have a sum of $45$ and their difference is $13$ . If the larger number is $x$ and the smaller number is $y$ , and you solve the system $x + y = 45$ and $x - y = 13$ using substitution by solving the first equation for $y$ , what is the value of the larger number?

$x = 31$ , found by substituting $y = 45 - x$ and solving $x - (45 - x) = 13$
$x = 29$ , found by substituting $y = 45 - x$ and solving $x - (45 - x) = 13$ (correct answer)
$x = 32$ , found by substituting $y = 45 - x$ and solving $x - (45 - x) = 13$
$x = 16$ , found by substituting $y = 45 - x$ and solving $x - (45 - x) = 13$

Explanation: When you encounter a system of linear equations like this, you're looking at a classic substitution problem where you solve one equation for a variable, then substitute that expression into the other equation. Starting with the system

x + y = 45

and

x - y = 13

, you first solve the equation

x + y = 45

for

y

y = 45 - x

. Now substitute this expression into the second equation:

x - (45 - x) = 13

. Distributing the negative sign:

x - 45 + x = 13

. Combining like terms:

2x - 45 = 13

. Adding 45 to both sides:

2x = 58

. Dividing by 2:

x = 29

. You can verify this by finding

y = 45 - 29 = 16

, and checking:

29 + 16 = 45

✓ and

29 - 16 = 13

✓. Looking at the wrong answers: Choice A gives

x = 31

, which would make

y = 14

, but

31 - 14 = 17 \neq 13

. Choice C suggests

x = 32

, making

y = 13

, but

32 - 13 = 19 \neq 13

. Choice D claims

x = 16

, which would actually make this the smaller number since

16 < 29

, contradicting the problem setup where

x

represents the larger number. The correct answer is B:

x = 29

. Remember: when solving systems by substitution, always verify your solution in both original equations. This catches arithmetic errors and ensures you haven't mixed up which variable represents which quantity.

Question 4

The system $2x + 3y = 12$ and $4x + 6y = k$ is being solved using substitution. From the first equation, $x = 6 - \frac{3y}{2}$ . When this expression is substituted into the second equation, what must be true about the value of $k$ for the system to have infinitely many solutions?

$k = 24$ , because substitution yields $24 = k$ , making the equations equivalent and consistent (correct answer)
$k = 12$ , because substitution yields $12 = k$ , making the equations equivalent and consistent
$k = 18$ , because substitution yields $18 = k$ , making the equations equivalent and consistent
$k = 36$ , because substitution yields $36 = k$ , making the equations equivalent and consistent

Explanation: Substituting

x = 6 - \frac{3y}{2}

into

4x + 6y = k

4(6 - \frac{3y}{2}) + 6y = k

, which gives

24 - 6y + 6y = k

, so

24 = k

. When

k = 24

, the second equation becomes

4x + 6y = 24

, which is exactly twice the first equation, making them equivalent. Choice B incorrectly copies the constant from the first equation. Choice C results from arithmetic errors in the substitution. Choice D incorrectly multiplies the first equation by 3 instead of recognizing the factor of 2 relationship.

Question 5

The system $3x - 4y = 7$ and $6x - 8y = 14$ is being solved using substitution. After solving the first equation for $x$ to get $x = \frac{7 + 4y}{3}$ and substituting this into the second equation, what do you conclude about the system?

The system has exactly one solution because substitution leads to a solvable equation for $y$ , then back-substitution gives $x$
The system has no solution because substitution leads to the contradiction $14 = 28$ , indicating inconsistent equations
The system has infinitely many solutions because substitution leads to the identity $14 = 14$ , indicating dependent equations (correct answer)
The system has no solution because substitution leads to the contradiction $7 = 14$ , indicating inconsistent equations

Explanation: When solving systems of linear equations, you need to recognize whether the equations are independent (one solution), inconsistent (no solution), or dependent (infinitely many solutions). This recognition becomes clear during the substitution process. Let's complete the substitution. Starting with

x = \frac{7 + 4y}{3}

, substitute this into the second equation:

6x - 8y = 14

6 \cdot \frac{7 + 4y}{3} - 8y = 14

2(7 + 4y) - 8y = 14

14 + 8y - 8y = 14

14 = 14

This identity tells us the equations are dependent—they're actually the same line written in different forms. Notice that if you multiply the first equation by 2, you get:

2(3x - 4y) = 2(7)

, which gives

6x - 8y = 14

—exactly the second equation. Option C is correct because the substitution yields

14 = 14

, indicating the system has infinitely many solutions along the line

3x - 4y = 7

. Option A is wrong because we don't get a solvable equation for

y

—the variable cancels out completely. Option B incorrectly states we get

14 = 28

, which isn't what happens in our calculation. Option D also gives a wrong contradiction (

7 = 14

) and misses that dependent equations mean infinite solutions, not no solutions. Remember: when variables cancel and you get a true statement like

14 = 14

, the system has infinitely many solutions. A false statement would mean no solution.

Question 6

A bakery makes two types of cookies. The cost equation for chocolate chip cookies is $C_1 = 2x + 15$ , and for sugar cookies is $C_2 = 3x + 5$ , where $x$ represents dozens of cookies and $C$ represents cost in dollars. To find when both types cost the same amount, you set up the equation $2x + 15 = 3x + 5$ . Solving this by substitution method (treating it as the system $y = 2x + 15$ and $y = 3x + 5$ ), how many dozens of cookies result in equal costs?

$x = 20$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$
$x = 5$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$
$x = 15$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$
$x = 10$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$ (correct answer)

Explanation: When you encounter cost equations that need to be set equal, you're looking for the intersection point where both expressions have the same value. This is a classic system of equations problem disguised as a business scenario. To find when both cookie types cost the same, you correctly set up

2x + 15 = 3x + 5

. Now you need to solve for

x

. Subtract

2x

from both sides:

15 = x + 5

. Then subtract

5

from both sides:

10 = x

. So

x = 10

dozens of cookies. You can verify this: when

x = 10

, chocolate chip cookies cost

C_1 = 2(10) + 15 = 35

dollars, and sugar cookies cost

C_2 = 3(10) + 5 = 35

dollars. Both equal

35

, confirming our answer. Choice A gives

x = 20

, but substituting shows

C_1 = 55

while

C_2 = 65

— these aren't equal. Choice B suggests

x = 5

, yielding

C_1 = 25

and

C_2 = 20

— again, not equal. Choice C proposes

x = 15

, giving

C_1 = 45

and

C_2 = 50

— still unequal. Notice that all answer choices mention the same setup equation, which is correct. The error in the wrong answers lies in the algebraic solution, not the problem setup. Study tip: Always verify your solution by substituting back into both original equations. If the outputs don't match exactly, you've made an algebraic error and should re-solve.

Question 7

A clothing store sells shirts and pants. Last week, $5$ shirts and $3$ pants cost $139, while $ $2$ $shirts and$ $7$ $ pants cost $146. Using substitution to solve for the individual prices, if you first solve for the price of a shirt from the first equation, what equation do you get when you substitute into the second equation?

$2\left(\frac{139 - 5p}{3}\right) + 7p = 146$ , which simplifies to $\frac{278 - 10p + 21p}{3} = 146$
$2\left(\frac{139 + 3p}{5}\right) + 7p = 146$ , which simplifies to $\frac{278 + 6p + 35p}{5} = 146$
$2\left(\frac{139 - 3p}{5}\right) + 7p = 146$ , which simplifies to $\frac{278 - 6p + 35p}{5} = 146$ (correct answer)
$2s + 7\left(\frac{139 - 5s}{3}\right) = 146$ , which simplifies to $\frac{6s + 973 - 35s}{3} = 146$

Explanation: When you encounter a system of equations problem asking about substitution, you need to carefully track which variable you're solving for and how you manipulate the equations. Let's set up the system first. If

s

= price of a shirt and

p

= price of pants, then:

First equation: $5s + 3p = 139$
Second equation: $2s + 7p = 146$

The question asks you to solve for the shirt price from the first equation. Isolating

s

5s = 139 - 3p

s = \frac{139 - 3p}{5}

Now substitute this expression for

s

into the second equation:

2\left(\frac{139 - 3p}{5}\right) + 7p = 146

To simplify, multiply the first term by 2 and combine with

7p

. Getting a common denominator of 5:

\frac{2(139 - 3p) + 35p}{5} = 146

\frac{278 - 6p + 35p}{5} = 146

This matches answer choice C exactly. Answer A incorrectly solves for

s

in terms of

p

from the wrong rearrangement, getting

s = \frac{139 - 5p}{3}

. Answer B makes an algebra error, changing the sign to get

s = \frac{139 + 3p}{5}

. Answer D correctly identifies the substitution process but solves for

p

instead of

s

from the first equation. Study tip: When doing substitution problems, always double-check which variable you're asked to eliminate first, and be extra careful with your algebra when isolating variables—sign errors are common pitfalls.

Question 8

A tutoring center charges a monthly membership fee plus an hourly rate for sessions. Maria paid $180 for $ $8$ $hours of tutoring in one month, while David paid $240 for $$ 12$$ hours of tutoring in the same month. Using the substitution method to solve the system of equations, what is the hourly rate for tutoring sessions?

$12 per hour with a $84 monthly membership fee, verified by checking both original equations
$15 per hour with a $60 monthly membership fee, verified by checking both original equations (correct answer)
$18 per hour with a $36 monthly membership fee, verified by checking both original equations
$20 per hour with a $20 monthly membership fee, verified by checking both original equations

Explanation: Let

m

= monthly membership fee and

h

= hourly rate. The system is:

m + 8h = 180

and

m + 12h = 240

. From the first equation:

m = 180 - 8h

. Substituting into the second equation:

(180 - 8h) + 12h = 240

, which simplifies to

180 + 4h = 240

, so

4h = 60

and

h = 15

. Then

m = 180 - 8(15) = 60

. Choice A incorrectly uses elimination rather than substitution steps. Choice C represents an arithmetic error in the substitution process. Choice D results from incorrectly setting up the original equations.

Question 9

A concert venue sells general admission and VIP tickets. On Friday, they sold $120$ general admission tickets and $80$ VIP tickets for a total revenue of $3200. On Saturday, they sold $ $150$ $general admission tickets and$ $60$ $ VIP tickets for a total revenue of $3000. Using substitution to find the price of each ticket type, what is the cost of a VIP ticket?

$25, since the general admission ticket costs $15 and verification shows the system is satisfied (correct answer)
$20, since the general admission ticket costs $20 and verification shows the system is satisfied
$30, since the general admission ticket costs $10 and verification shows the system is satisfied
$35, since the general admission ticket costs $5 and verification shows the system is satisfied

Explanation: Let

g

= general admission price and

v

= VIP price. The system is:

120g + 80v = 3200

and

150g + 60v = 3000

. From the first equation:

g = \frac{3200 - 80v}{120} = \frac{80(40 - v)}{120} = \frac{2(40 - v)}{3}

. Substituting into the second equation and solving gives

v = 25

and

g = 15

. Choice B results from arithmetic errors in the substitution algebra. Choice C comes from incorrectly simplifying the first equation. Choice D represents errors in both the substitution step and subsequent solving.