College Algebra Quiz: Solve Systems Of Linear Equations Substitution

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.

Question 1

1. $x = 30 - 2y$, obtained by multiplying the first equation by 6 and solving for $x$
2. $x = 5 - \frac{y}{3}$, obtained by multiplying the first equation by 2 and solving for $x$
3. $x = 15 - y$, obtained by multiplying the first equation by 3 and solving for $x$
4. $x = 10 - \frac{2y}{3}$, 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

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

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

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

1. The system has exactly one solution because substitution leads to a solvable equation for $y$, then back-substitution gives $x$
2. The system has no solution because substitution leads to the contradiction $14 = 28$, indicating inconsistent equations
3. The system has infinitely many solutions because substitution leads to the identity $14 = 14$, indicating dependent equations (correct answer)
4. 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

1. $x = 20$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$
2. $x = 5$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$
3. $x = 15$ dozens, because substituting the first equation into the second gives $2x + 15 = 3x + 5$
4. $x = 10$ 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

1. $2\left(\frac{139 - 5p}{3}\right) + 7p = 146$, which simplifies to $\frac{278 - 10p + 21p}{3} = 146$
2. $2\left(\frac{139 + 3p}{5}\right) + 7p = 146$, which simplifies to $\frac{278 + 6p + 35p}{5} = 146$
3. $2\left(\frac{139 - 3p}{5}\right) + 7p = 146$, which simplifies to $\frac{278 - 6p + 35p}{5} = 146$ (correct answer)
4. $2s + 7\left(\frac{139 - 5s}{3}\right) = 146$, 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: $2s + 7p = 146$

Question 8

1. $12 per hour with a$84 monthly membership fee, verified by checking both original equations
2. $15 per hour with a$60 monthly membership fee, verified by checking both original equations (correct answer)
3. $18 per hour with a$36 monthly membership fee, verified by checking both original equations
4. $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

1. $25, since the general admission ticket costs$15 and verification shows the system is satisfied (correct answer)
2. $20, since the general admission ticket costs$20 and verification shows the system is satisfied
3. $30, since the general admission ticket costs$10 and verification shows the system is satisfied
4. $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.