College Algebra Quiz: Solve Systems Of Linear Equations Elimination

What this quiz covers

This quiz focuses on Solve Systems Of Linear Equations Elimination, 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. Multiply the first equation by 2 and add to the second equation unchanged
2. Multiply the first equation by -2 and add to the second equation unchanged (correct answer)
3. Multiply the first equation by 4 and the second equation by 2
4. Multiply the first equation by 5 and the second equation by 3

Explanation: To eliminate $x$, we need the coefficients of $x$ to be opposites. The first equation has coefficient 2 for $x$, and the second has coefficient 4. If we multiply the first equation by -2, we get $-4x + 6y = -14$. Adding this to $4x + 5y = 1$ gives $(-4x + 6y) + (4x + 5y) = -14 + 1$, which simplifies to $11y = -13$, eliminating $x$. Choice A gives $4x - 6y = 14$ plus $4x + 5y = 1$, resulting in $8x - y = 15$ (doesn't eliminate $x$). Choice C creates $8x - 12y = 28$ and $8x + 10y = 2$ (coefficients are the same, not opposites). Choice D creates $10x - 15y = 35$ and $12x + 15y = 3$ (eliminates $y$, not $x$).

Question 2

1. The system has solution $(3, 0)$ since the variables were eliminated successfully
2. The system has no solution since adding gives $0 = 6$, which is impossible (correct answer)
3. The system has infinitely many solutions since both variables were eliminated
4. The elimination was performed incorrectly and should be repeated with different multipliers

Explanation: When we add the equations $18x - 12y = 54$ and $-18x + 12y = -48$, we get $0x + 0y = 54 + (-48) = 6$, or simply $0 = 6$. This is a contradiction, which means the system has no solution (the lines are parallel but distinct). Choice A is wrong because we don't get a unique solution when both variables are eliminated. Choice C is incorrect because infinitely many solutions occurs when elimination gives $0 = 0$ (a true statement), not $0 = 6$. Choice D is wrong because the elimination was performed correctly; the contradiction reveals that the system is inconsistent.

Question 3

1. Substitute $x = \frac{59}{29}$ into either original equation and solve for $y$ (correct answer)
2. Use back-substitution with $x = 29$ in the equation $5x + 2y = 13$
3. Set up a new elimination to solve for $y$ since the current method failed
4. Substitute $x = \frac{91}{35}$ into the first modified equation to find $y$

Explanation: Adding $35x + 14y = 91$ and $-6x - 14y = -32$ gives $29x = 59$, so $x = \frac{59}{29}$. To find $y$, we substitute this value into either original equation. Using $5x + 2y = 13$: $5(\frac{59}{29}) + 2y = 13$, which gives $\frac{295}{29} + 2y = 13$, so $2y = 13 - \frac{295}{29} = \frac{377-295}{29} = \frac{82}{29}$, thus $y = \frac{41}{29}$. Choice B uses $x = 29$, which is incorrect (should be $\frac{59}{29}$). Choice C is wrong because elimination worked perfectly. Choice D uses an incorrect value $x = \frac{91}{35}$ and suggests using a modified equation rather than an original one.

Question 4

1. Multiply the first equation by -3 and the second equation by 2, then add the results
2. Multiply the first equation by -2 and the second equation by -5, then subtract the results
3. Multiply the first equation by 2 and the second equation by 5, then subtract the results
4. Multiply the first equation by 2 and the second equation by 5, then add the results (correct answer)

Explanation: When solving systems of equations by elimination, your goal is to make the coefficients of one variable opposites so they cancel when you add the equations together. Since you want to eliminate $y$, you need to make the $y$-coefficients in both equations opposites of each other. Looking at the system $2x + 5y = 19$ and $3x - 2y = 4$, the $y$-coefficients are $+5$ and $-2$. To make these opposites, you need a common multiple. The least common multiple of 5 and 2 is 10. To get $+10$ and $-10$, multiply the first equation by 2 (giving $+10y$) and the second equation by 5 (giving $-10y$). When you add these results, the $y$-terms will cancel: $+10y + (-10y) = 0$. Choice D correctly identifies these multipliers and the addition operation. Multiplying gives you $4x + 10y = 38$ and $15x - 10y = 20$. Adding these eliminates $y$ and yields $19x = 58$. Choice A uses incorrect multipliers that would eliminate $x$ instead of $y$. Choice B uses the wrong operation—when you have opposite coefficients, you add rather than subtract. Choice C also uses subtraction incorrectly; after applying the multipliers, you'd have $+10y$ in both equations, so subtracting would give $+20y$, not elimination. Study tip: For elimination, find the LCM of the coefficients you want to eliminate, then choose multipliers that create opposite signs. Always add the equations when the target coefficients have opposite signs.

Question 5

1. Eliminate $x$ by multiplying the first equation by 6 and the second by -3, then adding
2. Eliminate $y$ by multiplying the first equation by 2 and the second by 7, then adding
3. Eliminate $x$ by multiplying the first equation by -2 and adding to the second unchanged (correct answer)
4. Eliminate $y$ by multiplying the second equation by $\frac{7}{2}$ and adding to the first unchanged

Explanation: When solving systems of linear equations by elimination, you want to make the coefficients of one variable opposites so they cancel when you add the equations. The most efficient approach minimizes the complexity of multiplication operations. Let's examine the coefficients in our system: $3x - 7y = -4$ and $6x + 2y = 16$. Notice that the $x$-coefficients are 3 and 6, where 6 is exactly twice 3. This means you can eliminate $x$ by multiplying the first equation by -2, giving you $-6x + 14y = 8$. When you add this to the unchanged second equation $6x + 2y = 16$, the $x$-terms cancel perfectly: $16y = 24$, so $y = \frac{3}{2}$. This approach requires only one simple multiplication by -2. Option A involves multiplying by 6 and -3, creating much larger numbers (coefficients of ±18) that are harder to work with. Option B requires multiplying the first equation by 2 and the second by 7, again creating larger coefficients (14 and 49) and more complex arithmetic. Option D involves multiplying by the fraction $\frac{7}{2}$, which introduces fractional coefficients throughout your work—significantly more cumbersome than working with the integer -2. The key insight is that option C takes advantage of the existing relationship between coefficients (6 = 2 × 3) and uses the smallest possible multiplier. Strategy tip: Before choosing your elimination approach, scan both equations for coefficients that are already multiples of each other. This often reveals the path requiring the least arithmetic complexity.

Question 6

1. The solution satisfies both equations, confirming that $(2, 3)$ is indeed correct despite the elimination error
2. The solution satisfies the second equation but not the first, suggesting the wrong variable was eliminated
3. The solution satisfies the first equation but not the second, indicating an arithmetic error in elimination (correct answer)
4. The solution satisfies neither equation, indicating multiple errors in the elimination process

Explanation: When solving systems of equations by elimination, checking your solution in the original equations is crucial because it reveals where errors occurred during the process. This verification step helps you identify whether mistakes happened in specific calculations or variable eliminations. Let's check the proposed solution $(2, 3)$ in both original equations. For the first equation $7x + 4y = 26$: substituting gives us $7(2) + 4(3) = 14 + 12 = 26$. This works perfectly. For the second equation $5x - 3y = 13$: substituting gives us $5(2) - 3(3) = 10 - 9 = 1 \neq 13$. This doesn't work. Since $(2, 3)$ satisfies the first equation but not the second, the answer is C. This pattern indicates an arithmetic error occurred specifically when working with the second equation during elimination. Choice A is wrong because the solution doesn't satisfy both equations—it fails the second one. Choice B incorrectly suggests the solution satisfies only the second equation, when it's actually the opposite. Choice D is wrong because the solution does satisfy the first equation, so there wasn't a complete failure. Study tip: Always check your solutions in both original equations, not just one. When a solution satisfies one equation but not the other, you know your error occurred in calculations involving the equation that fails. This diagnostic approach helps you pinpoint exactly where to look for your mistake, making it easier to correct your work.

Question 7

1. Multiply the first equation by 6 and the second equation by 12 to clear all denominators (correct answer)
2. Multiply the first equation by 3 and the second equation by 4 to eliminate the largest denominators
3. Multiply both equations by their respective least common denominators of the coefficients
4. Convert all fractions to decimals before proceeding with standard elimination

Explanation: To clear fractions, we multiply each equation by the LCD of all denominators in that equation. For the first equation $\frac{1}{2}x + \frac{1}{3}y = 4$, the denominators are 2 and 3, so LCD = 6. Multiplying by 6: $3x + 2y = 24$. For the second equation $\frac{2}{3}x - \frac{1}{4}y = 1$, the denominators are 3 and 4, so LCD = 12. Multiplying by 12: $8x - 3y = 12$. Choice B doesn't clear all fractions (multiplying first by 3 gives $\frac{3}{2}x + y = 12$, still has a fraction). Choice C is vague and doesn't specify the correct LCD values. Choice D introduces rounding errors and is unnecessarily complicated compared to clearing fractions.