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slamgirl [31]
3 years ago
6

Solve the system of linear operations

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

\boxed{(-2,1)}

Step-by-step explanation:

\left \{ {{5x+2y=-8} \atop {x+4y=2}} \right.

I'll be solving this system of equations using the elimination method since the x and y values are neatly lined up.

I want to get a pair of x's or y's that cancel out, and it looks like the easiest way to start would be by multiplying the first equation by -2 (the y's will cancel).

I chose to multiply the first equation by -2 instead of multiplying the second equation by 5 because -2 is a smaller number and easier to multiply by.

-2\times(5x+2y=-8)

Distribute -2 inside the parentheses. Now you've got:

\left \{ {{-10x-4y=16} \atop {x+4y=2}} \right.

Add up the equations from top to bottom.

-10x plus x is -9x, the -4y and 4y cancel out, and 16 plus 2 is 18. Make this one single equation.

-9x=18

Divide both sides by -9.

x=-2

Substitute this value of x into the second equation (less to do with the x since it has no coefficient which means no multiplying).

(-2)+4y=2

Add 2 to both sides.

4y=4

Divide both sides by 4.

y=1

The final answer is x=-2, ~y=1.

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