Solve the system of linear operations
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Answer:
[tex]\boxed{(-2,1)}[/tex]
Step-by-step explanation:
[tex]\left \{ {{5x+2y=-8} \atop {x+4y=2}} \right.[/tex]
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.
[tex]-2\times(5x+2y=-8)[/tex]
Distribute -2 inside the parentheses. Now you've got:
[tex]\left \{ {{-10x-4y=16} \atop {x+4y=2}} \right.[/tex]
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.
[tex]-9x=18[/tex]
Divide both sides by -9.
[tex]x=-2[/tex]
Substitute this value of x into the second equation (less to do with the x since it has no coefficient which means no multiplying).
[tex](-2)+4y=2[/tex]
Add 2 to both sides.
[tex]4y=4[/tex]
Divide both sides by 4.
[tex]y=1[/tex]
The final answer is [tex]x=-2, ~y=1[/tex].