If [tex]y=x+6[/tex] and [tex]y=-2x-3[/tex], then you can replace [tex]y[/tex] in either equation with the equivalent expression given by the other equation. In this case, you have
[tex]x+6=-2x-3[/tex]
and you can solve for [tex]x[/tex].
[tex]x+2x+6-6=-2x+2x-3+6\implies3x=3\implies\dfrac{3x}3=\dfrac33\impliesx=1[/tex]
Then [tex]y=1+6=7[/tex].
For the second system, you can write
[tex]5x-y=-5\implies 5x-5x-y=-5-5x\implies-y=-5-5x[/tex]
[tex]\implies-(-y)=-(-5-5x)\implies y=5+5x[/tex]
Then substitute this into the second equation and you get
[tex]3x-6(5+5x)=24\implies3x-30-30x=24\implies-27x-30+30=24+30[/tex]
[tex]\implies-27x=54\implies x=2[/tex]
so that
[tex]y=5+5(2)=15[/tex]
