Answered

Trevor solved the system of equations below. What mistake did he make in his work? 2x + y = 5 x − 2y = 10 y = 5 − 2x x − 2(5 − 2x) = 10 x − 10 + 4x = 10 5x − 10 = 10 5x = 0 x = 0 2(0) + y = 5 y = 5

AHe should have substituted 5 + 2x
BHe combined like terms incorrectly, it should have been 4x instead of 5x
CHe subtracted 10 from the right side instead of adding 10 to the right side
DHe made no mistake

Verified Answer :

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\(2x+y=5 \\ x-2y=10 \\ \\ y=5-2x \\ x-2(5-2x)=10 \\ x-10+4x=10 \\ 5x-10=10 \\ \boxed{5x=0} \Leftarrow \hbox{the mistake, it should be 5x=20} \\ x=0 \\ \\ 2(0)+y=5 \\ y=5\)

If 2x+y=5, then y=5-2x, so he substituted 5-2x correctly.
x+4x=5x, so he combined like terms correctly.
If 5x-10=10, then 5x=10+10 -> 5x=20, so he subtracted 10 from the right side instead of adding 10 to the right side.

The answer is C.

Here's the correct solution:
\(2x+y=5 \\ x-2y=10 \\ \\ y=5-2x \\ x-2(5-2x)=10 \\ x-10+4x=10 \\ 5x-10=10 \\ 5x=10+10 \\ 5x=20 \\ x=\frac{20}{5} \\ x=4 \\ \\ y=5-2x \\ y=5 - 2 \times 4 \\ y=5-8 \\ y=-3 \\ \\ (x,y)=(4,-3)\)

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