Mathematics
Mathematics, 02.04.2021 19:20, fdzgema17

Kathleen successfully solved the system of equations 4x+3y=15 and βˆ’5xβˆ’2y=βˆ’24
by multiplying each equation by a constant and adding the two equations together to eliminate the x
-terms. She then solved for y
and substituted the value of y
back into one of the equations and solved for x
. What could she have multiplied each equation by, and what was her final answer?

A. She could have multiplied the first equation by 4 and the second equation by 5, getting a final answer of x=βˆ’3
and y=6.

B. equation by 5
and the second equation by 4, getting a final answer of x=βˆ’3 and y=6.

C. She could have multiplied the first equation by 4
and the second equation by 5, getting a final answer of x=6 and y=βˆ’3.

D. She could have multiplied the first equation by 5
and the second equation by 4, getting a final answer of x=6 and y=βˆ’3.

answer
Answers: 1

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Do you know the correct answer?
Kathleen successfully solved the system of equations 4x+3y=15 and βˆ’5xβˆ’2y=βˆ’24
by multiplying e...

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