Solving Systems by Elimination - Math 1

After applying elimination to a system of linear equations, a student obtains the equation $0 = 0$ . The student concludes the system has infinitely many solutions. Under what condition is this conclusion correct?
Answer: The conclusion is correct only if the original equations are scalar multiples of each other Getting $0 = 0$ during elimination means one equation is a scalar multiple of the other, making them represent the same line (infinitely many solutions). However, if the equations had been inconsistent (parallel lines), elimination would yield $0 = c$ where $c \neq 0$ . Choice A is incorrect because $0 = 0$ specifically indicates dependent equations. Choice C is imprecise—they must be the same line, not just have the same slope. Choice D is wrong; $0 = 0$ is a valid mathematical result indicating dependency.
When solving $7x - 2y = 13$ and $3x + 5y = 4$ by elimination, a student decides to eliminate $x$ first. They multiply the first equation by $3$ and the second by $7$ . What should be their next step to complete the elimination correctly?
Answer: Recognize that both equations now have $21x$ , so either subtraction order will eliminate $x$ successfully After multiplication, we have $21x - 6y = 39$ and $21x + 35y = 28$ . Since both have the same $x$ -coefficient ( $21x$ ), subtracting either equation from the other will eliminate $x$ . Choice B gives: $21x + 35y - 21x + 6y = 28 - 39$ , so $41y = -11$ . Choice C gives: $21x - 6y - 21x - 35y = 39 - 28$ , so $-41y = 11$ . Both lead to the same solution: $y = -\frac{11}{41}$ . Choice A is incorrect because adding keeps both variables. Choices B and C are both valid approaches, making D the best answer since it recognizes that either subtraction order works.
To solve $\frac{x}{2} + \frac{y}{3} = 4$ and $\frac{x}{4} - \frac{y}{6} = 1$ by elimination, what is the most efficient first step?
Answer: Clear denominators in each equation separately, then apply standard elimination techniques to the resulting system The most systematic approach is to first clear denominators in each equation. Multiply the first equation by 6: $3x + 2y = 24$ . Multiply the second equation by 12: $3x - 2y = 12$ . Now the system $3x + 2y = 24$ and $3x - 2y = 12$ can be easily solved by adding to eliminate $y$ : $6x = 36$ , so $x = 6$ . Choice A uses different multipliers (6 and 12) which is unnecessarily complex. Choice B attempts elimination before clearing fractions, which leads to more complex fractional arithmetic. Choice D also tries to eliminate before simplifying the fractions, making the process more error-prone.
A student attempts to solve the system $3x + 2y = 14$ and $5x - 4y = 2$ by elimination. After multiplying the first equation by 2, they obtain $6x + 4y = 28$ . When they add this to the second equation, what is the resulting equation in one variable?
Answer: $11x = 30$ After multiplying the first equation by 2: $6x + 4y = 28$ . Adding this to the second equation $5x - 4y = 2$ : $(6x + 4y) + (5x - 4y) = 28 + 2$ , which gives $11x = 30$ . Choice B results from adding incorrectly (28 - 2 instead of 28 + 2). Choice C omits the coefficient 11. Choice D results from subtracting the equations instead of adding them.
When solving the system $5x - 2y = 13$ and $3x + 7y = 1$ by elimination, a student decides to eliminate $y$ first. What is the smallest positive integer that the student could multiply the first equation by to achieve this elimination?
Answer: 7 To eliminate $y$ , the coefficients of $y$ must be opposites. The first equation has $-2y$ and the second has $+7y$ . The LCM of 2 and 7 is 14. To get $+14y$ in the first equation, multiply by 7: $35x - 14y = 91$ . To get $-14y$ in the second equation, multiply by 2: $6x - 14y = 2$ . The smallest positive multiplier for the first equation is 7. Choice A (2) doesn't create opposite coefficients. Choice C (14) works but isn't the smallest. Choice D (5) is irrelevant to the $y$ -coefficients.
To solve the system $2x + 3y = 8$ and $4x - y = 5$ by elimination, which of the following first steps would require the fewest arithmetic operations to eliminate one variable?
Answer: Multiply the second equation by $3$ and add it to the first equation to eliminate $y$ To eliminate $y$ , multiply the second equation by $3$ : $12x - 3y = 15$ . Adding this to $2x + 3y = 8$ gives $14x = 23$ , requiring only one multiplication and one addition. Choice A gives $4x + 6y = 16$ and subtracting $4x - y = 5$ yields $7y = 11$ , but subtraction is more error-prone. Choice C gives $-4x - 6y = -16$ and adding $4x - y = 5$ yields $-7y = -11$ , involving negative coefficients. Choice D requires multiplying by $2$ then subtracting, which is less efficient than choice B.
Two students solve the system $3x + 4y = 11$ and $2x - y = 1$ by elimination. Student A multiplies the second equation by 4 before adding. Student B multiplies the first equation by 2 and the second by -3 before adding. Which statement is true?
Answer: Both students will eliminate a variable completely on the first addition Student A: $3x + 4y = 11$ and $8x - 4y = 4$ . Adding eliminates $y$ : $11x = 15$ . Student B: $6x + 8y = 22$ and $-6x + 3y = -3$ . Adding eliminates $x$ : $11y = 19$ . Both methods successfully eliminate one variable. Choices A and B incorrectly suggest only one method works. Choice D incorrectly suggests neither works.
The system $\frac{1}{2}x + \frac{1}{3}y = 4$ and $\frac{3}{4}x - \frac{1}{6}y = 1$ is to be solved by elimination. Before applying elimination, what is the most efficient first step?
Answer: Multiply the first equation by 6 and the second equation by 12 When solving systems of linear equations with fractions using elimination, your goal is to clear denominators efficiently while setting up coefficients that will eliminate easily. The key is finding the least common denominator (LCD) for each equation separately, which will give you the cleanest integer coefficients to work with. For the first equation $\frac{1}{2}x + \frac{1}{3}y = 4$ , the denominators are 2 and 3, so the LCD is 6. For the second equation $\frac{3}{4}x - \frac{1}{6}y = 1$ , the denominators are 4 and 6, so the LCD is 12. Multiplying by these values gives you: - First equation × 6: $3x + 2y = 24$ - Second equation × 12: $9x - 2y = 12$ Notice that the y-coefficients are now +2 and -2, which will eliminate perfectly when you add the equations. Option A only clears denominators from the first equation, leaving fractions in your work. Option B is too vague—you need to identify the specific LCDs, which are 6 and 12 respectively. Option C converts to decimals unnecessarily, creating messier arithmetic when integers work perfectly well. Option D correctly identifies that multiplying the first equation by 6 and the second by 12 will clear all fractions and create the most efficient setup for elimination. Study tip: Always clear fractions first by multiplying each equation by its LCD, then look for elimination opportunities. This two-step approach—clear denominators, then eliminate—will save you time and reduce errors.
Consider the system $kx + 3y = 9$ and $2x + 6y = 18$ . For what value of $k$ will elimination by adding appropriate multiples of these equations result in the identity $0 = 0$ ?
Answer: $k = 1$ , because this makes the first equation become $x + 3y = 9$ , which is exactly half the second For elimination to yield $0 = 0$ , the equations must be equivalent (one is a scalar multiple of the other). The second equation $2x + 6y = 18$ can be divided by 2 to get $x + 3y = 9$ . For the first equation $kx + 3y = 9$ to be equivalent, we need $k = 1$ . Then both equations become $x + 3y = 9$ , and subtracting gives $0 = 0$ . Choice B makes the first equation $2x + 3y = 9$ , which is not proportional to $2x + 6y = 18$ . Choice C gives $6x + 3y = 9$ , also not proportional. Choice D gives $3x + 3y = 9$ , which is not equivalent to the second equation.
A student solving $3x + 4y = 20$ and $2x - 4y = -5$ writes: "Adding the equations: $5x + 0y = 15$ , so $x = 3$ . Substituting into the first equation: $3(3) + 4y = 20$ , so $9 + 4y = 20$ , giving $4y = 11$ , thus $y = \frac{11}{4}$ . Checking in the second equation: $2(3) - 4(\frac{11}{4}) = 6 - 11 = -5$ ✓." What is the most significant issue with this solution?
Answer: All steps are mathematically correct and the solution $(3, \frac{11}{4})$ satisfies both original equations Let's verify each step: Adding $3x + 4y = 20$ and $2x - 4y = -5$ gives $5x = 15$ , so $x = 3$ ✓. Substituting: $3(3) + 4y = 20$ gives $9 + 4y = 20$ , so $4y = 11$ and $y = \frac{11}{4}$ ✓. Checking: $2(3) - 4(\frac{11}{4}) = 6 - 11 = -5$ ✓. Choice A is wrong because $20 + (-5) = 15$ , not 25. Choice B is wrong because $\frac{11}{4} = 2.75$ , which is the same value. Choice C is wrong because the calculation $6 - 11 = -5$ matches the right side of the second equation. All work is correct.
A system $Ax + By = C$ and $Dx + Ey = F$ is solved by elimination. After multiplying the first equation by $E$ and the second by $-B$ , the resulting equation is $x(AE - BD) = CE - BF$ . If $AE - BD \neq 0$ , what does this tell us about the original system?
Answer: The system has exactly one solution, and $y$ can be found by substituting $x = \frac{CE - BF}{AE - BD}$ back into either equation When solving a system of linear equations by elimination, the condition $AE - BD \neq 0$ reveals crucial information about the system's solution set. This expression is actually the determinant of the coefficient matrix, and when it's non-zero, it guarantees the system has exactly one unique solution. The elimination process here successfully isolated $x$ , giving us $x = \frac{CE - BF}{AE - BD}$ . Since $AE - BD \neq 0$ , we can divide by this expression to get a specific value for $x$ . Once you have this $x$ -value, you can substitute it back into either original equation to solve for $y$ , yielding a unique ordered pair solution. Choice A is incorrect because there's no contradiction here – the elimination worked perfectly and gave us a valid expression for $x$ . Choice C confuses the determinant condition: when $AE - BD \neq 0$ , the equations are NOT scalar multiples of each other, which is why we get a unique solution rather than infinitely many. Choice D misses the point – since we can solve for $x$ regardless of whether $CE - BF = 0$ , the system is automatically consistent and has a solution. The key insight is that $AE - BD$ is the determinant of your coefficient matrix. When this determinant is non-zero, you're guaranteed exactly one solution. When you see elimination leading to a solvable equation for one variable (with no contradictions like " $0 = 5$ "), the system has a unique solution.
Consider solving $mx + ny = p$ and $rx + sy = t$ by elimination, where all variables represent nonzero constants. The process involves multiplying the first equation by $s$ and the second by $-n$ , then adding. Under what condition will this process fail to produce a solution?
Answer: When $ms = rn$ , because the $x$ -coefficients become equal and don't eliminate as intended When solving systems of linear equations by elimination, you're strategically manipulating coefficients to eliminate one variable. The key insight is understanding what happens when your elimination strategy doesn't work as planned. Let's trace through the given process. You start with: - $mx + ny = p$ - $rx + sy = t$ After multiplying the first equation by $s$ and the second by $-n$ , you get: - $smx + sny = sp$ - $-nrx - nsy = -nt$ Adding these equations: $(sm - nr)x + (sn - ns)y = sp - nt$ Notice that the $y$ -terms always eliminate completely since $sn - ns = 0$ . However, if $sm - nr = 0$ (which means $ms = rn$ ), then the $x$ -coefficient also becomes zero. You're left with $0x + 0y = sp - nt$ , or simply $0 = sp - nt$ . This eliminates both variables when you only intended to eliminate $y$ , making the process fail to solve for $x$ . Choice A incorrectly focuses on when $ps - tn = 0$ , but this would actually give you $0 = 0$ , indicating infinitely many solutions rather than process failure. Choice B states $ms + rn = 0$ , but the correct coefficient after elimination is $(ms - rn)$ , not $(ms + rn)$ . Choice D mentions $ns + sr = 0$ , but the $y$ -coefficients should eliminate regardless—that's the point of this elimination method. Remember: elimination fails when you accidentally eliminate both variables instead of just the intended one. Always check that your multiplication factors don't make both variable coefficients zero.
The system $4x + by = 12$ and $ax + 3y = 9$ has no solution when solved by elimination. If $a = 6$ , what must be true about $b$ ?
Answer: $b = 2$ , making the left sides of the equations proportional while the right sides are not For no solution, the coefficient ratios must be equal but the constant ratio must be different. With $a = 6$ , the system becomes $4x + by = 12$ and $6x + 3y = 9$ . For no solution: $\frac{4}{6} = \frac{b}{3}$ but $\frac{4}{6} \neq \frac{12}{9}$ . From the first condition: $\frac{2}{3} = \frac{b}{3}$ , so $b = 2$ . Check the constant ratio: $\frac{12}{9} = \frac{4}{3} \neq \frac{2}{3}$ ✓. This confirms no solution. Choice B gives wrong coefficient ratios. Choice C gives $\frac{4}{6} = \frac{2}{3}$ and $\frac{b}{3} = \frac{3/2}{3} = \frac{1}{2} \neq \frac{2}{3}$ . Choice D is incorrect because $b = 2$ gives no solution, not infinitely many.
A system of equations $px + qy = r$ and $sx + ty = u$ is solved by elimination. After appropriate multiplication, the equations become $tpx + qty = tr$ and $spx + qty = su$ . What can be concluded if $tp \neq sp$ ?
Answer: The system has exactly one solution since subtracting eliminates $y$ and leaves a solvable equation in $x$ Since $tp \neq sp$ , subtracting the second equation from the first gives $(tp - sp)x + 0y = tr - su$ , which simplifies to $(tp - sp)x = tr - su$ . Since $tp - sp \neq 0$ , this equation has exactly one solution: $x = \frac{tr - su}{tp - sp}$ . This $x$ -value can then be substituted back to find a unique $y$ -value. Choice B is incorrect because different $x$ -coefficients with the same $y$ -coefficient actually allows elimination. Choice C is wrong because equal $y$ -coefficients enable elimination, not infinite solutions. Choice D is wrong because the condition $tp \neq sp$ alone guarantees a unique solution regardless of the relationship between $tr$ and $su$ .
The elimination method is applied to solve $mx + ny = p$ and $rx + sy = t$ . After one elimination step, the resulting equation is $0x + ky = c$ where $k \neq 0$ and $c \neq 0$ . What can be concluded about the relationship between $m$ , $n$ , $r$ , and $s$ ?
Answer: $ms = nr$ and the system has exactly one solution When you encounter elimination problems that result in equations like $0x + ky = c$ , you're dealing with a key insight about the coefficients of the original system and what this tells you about the solution. Let's think through what happened here. During elimination, you likely multiplied the first equation by some value and the second by another value to eliminate the $x$ terms. When the $x$ coefficient becomes zero after elimination, it means the $x$ coefficients in the original equations were proportional - specifically, $ms = nr$ . Since we get $0x + ky = c$ where $k \neq 0$ and $c \neq 0$ , we can solve directly: $y = \frac{c}{k}$ . With this specific $y$ -value, we can substitute back into either original equation to find a unique $x$ -value. This gives us exactly one solution to the system. Looking at the wrong answers: Choice A incorrectly states $ms \neq nr$ , but elimination only produces $0x$ when coefficients are proportional. Choice B makes the same coefficient error. Choice C correctly identifies $ms = nr$ but wrongly concludes no solutions exist - that would only happen if we got $0x + 0y = c$ with $c \neq 0$ . The correct answer is D: $ms = nr$ and the system has exactly one solution. Study tip: When elimination produces $0x + ky = c$ with $k \neq 0$ , you have proportional $x$ -coefficients and a unique solution. If you get $0x + 0y = c$ (with $c \neq 0$ ), then you have no solutions.
When solving the system $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{c} + \frac{y}{d} = 1$ by elimination (where $a, b, c, d$ are positive constants with $a \neq c$ and $b \neq d$ ), what is the most systematic approach?
Answer: Clear denominators by multiplying the first equation by $ab$ and the second by $cd$ , creating integer coefficients before elimination The most systematic approach is to clear denominators first. Multiplying $\frac{x}{a} + \frac{y}{b} = 1$ by $ab$ gives $bx + ay = ab$ . Multiplying $\frac{x}{c} + \frac{y}{d} = 1$ by $cd$ gives $dx + cy = cd$ . Now we have the system $bx + ay = ab$ and $dx + cy = cd$ with integer coefficients, making standard elimination techniques straightforward. Choice A over-complicates by using a common denominator for both equations. Choice B incorrectly describes cross-multiplication. Choice C attempts elimination before fully clearing fractions, which can lead to more complex arithmetic with mixed fractions.

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