To solve the system using substitution, first we need to transform one of the equations such that one of the variables is expressed as an equality in terms of the other variable(s).
[1] 3(x +5) - 2(y-1) = 6[1]3(x+5)−2(y−1)=6
[2] x + 4(y+3) = 13[2]x+4(y+3)=13
Let us isolate xx in equation [2][2]
[2] x + 4(y+3) = 13[2]x+4(y+3)=13
[2] => x = 13 - 4(y + 3)[2]⇒x=13−4(y+3)
[2] => x = 13 - 4y - 12[2]⇒x=13−4y−12
[2] => x = -4y + 1[2]⇒x=−4y+1
We then substitute xx in equation [1][1] with its equivalent we obtained in equation [2][2].
[1] 3(x + 5) - 2(y - 1) = 6[1]3(x+5)−2(y−1)=6
[1] => 3((-4y + 1) + 5) - 2(y-1) = 6[1]⇒3((−4y+1)+5)−2(y−1)=6
[1] => 3(-4y + 6) -2(y-1) = 6[1]⇒3(−4y+6)−2(y−1)=6
[1] => -12y + 18 -2y + 2 = 6[1]⇒−12y+18−2y+2=6
[1] => -14y = 6 - 18 - 2[1]⇒−14y=6−18−2
[1] => -14y = -14[1]⇒−14y=−14
[1] => y = 1[1]⇒y=1
To get xx, replace yy with its value in any of the equations above
x = -4y + 1x=−4y+1
x = -4(1) + 1x=−4(1)+1
x = -4 + 1x=−4+1
x = -3x=−3
