We are given two equations, one of which has an isolated variable [tex]x[/tex].
That screams to me that substitution would be a prefered strategy here, compared to elimination, although both work.
That means we'll be substituting our value of [tex]x[/tex], which is given as [tex]-y+3[/tex], into the first equation, [tex]15x+31y=-3[/tex].
[tex]15x+31y=-3[/tex]
[tex]x=-y+3[/tex]
[tex]15(-y+3)+31y=-3[/tex]
[tex]-15y+45+31y=-3[/tex]
[tex]16y=-48[/tex]
[tex]y=-3[/tex]
With this value, we can plug it back into either of the two equations to solve for [tex]x[/tex], I'll be substituting it back into the second equation, since it's easier.
[tex]x=-y+3[/tex]
[tex]y=-3[/tex]
[tex]x=-(-3)+3[/tex]
[tex]x=6[/tex]
So our solution is [tex](6,-3)[/tex], and to check we can plug it back into the first equation.
[tex]15x+31y=-3[/tex]
[tex]15(6)+31(-3)=-3[/tex]
[tex]90-93=-3[/tex]
Which is true, so our solution is correct.
Hope this helps!
