Question

How do you solve systems by substitution for 3x + 3y =-4 5x=20

Answer 1

$\left\{\begin{array}{ccc}3x+3y=-4\\5x=20&\ |divide\ both\ sides\ by\ 5\end{array}\right\\\\\left\{\begin{array}{ccc}3x+3y=-4\\x=4\end{array}\right\\\\substitute:\\\\3\cdot4+3y=-4\\12+3y=-4\\3y=-4-12\\3y=-16\ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\y=-\frac{16}{3}\\\\Solution:x=4\ and\ y=-\frac{16}{3}\to(4;-\frac{16}{3})$