Solve Systems of Linear Equations: Elimination - College Algebra

1. When solving $5x + 2y = 13$ and $3x + 7y = 16$ by elimination, a student multiplies the first equation by 7 and the second by -2 to get $35x + 14y = 91$ and $-6x - 14y = -32$. After adding these equations, what is the next step to find $y$?
   Answer: Substitute $x = \frac{59}{29}$ into either original equation and solve for $y$ 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.
2. A student solving $7x + 4y = 26$ and $5x - 3y = 13$ by elimination makes an error and gets the solution $(2, 3)$. When checking this solution in the original equations, what will they discover?
   Answer: The solution satisfies the first equation but not the second, indicating an arithmetic error in elimination 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.
3. To solve the system $2x - 3y = 7$ and $4x + 5y = 1$ by elimination, which of the following multiplications would eliminate $x$ when the resulting equations are added?
   Answer: Multiply the first equation by -2 and add to the second equation unchanged 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 $x$).
4. Consider the system $\frac{1}{2}x + \frac{1}{3}y = 4$ and $\frac{2}{3}x - \frac{1}{4}y = 1$. To eliminate fractions before using elimination, what should be the first step?
   Answer: Multiply the first equation by 6 and the second equation by 12 to clear all denominators 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.
5. In solving $3x - 7y = -4$ and $6x + 2y = 16$ by elimination, which approach requires the fewest arithmetic operations?
   Answer: Eliminate $x$ by multiplying the first equation by -2 and adding to the second unchanged 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.
6. A student attempts to solve $6x - 4y = 18$ and $9x - 6y = 24$ by elimination. After multiplying the first equation by 3 and the second by -2, they get $18x - 12y = 54$ and $-18x + 12y = -48$. What should they conclude?
   Answer: The system has no solution since adding gives $0 = 6$, which is impossible 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.
7. Using elimination to solve $2x + 5y = 19$ and $3x - 2y = 4$, a student decides to eliminate $y$ first. What is the most efficient choice of multipliers?
   Answer: Multiply the first equation by 2 and the second equation by 5, then add the results 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.

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