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].