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=-(16)/(3)\\\\Solution:x=4\ and\ y=-(16)/(3)\to(4;-(16)/(3))`