Question

How do you solve this system of equations algebraically with all steps showing? y=x^2+2x y=3x+20

Answer 1

`\bf \begin{cases} y=x^2+2x\\ y=3x+20 \end{cases}\implies \stackrel{y~~ =~~ y}{x^2+2x=3x+20} \\\\\\ x^2-1x-20=0\quad \begin{cases} -5+4=-1\\ -5\cdot 4=-20 \end{cases}\implies (x-5)(x+4)=0 \\\\\\ \begin{cases} x-5=0\implies &x=5\\ x+4=0\implies &x=-4 \end{cases}`


`\bf \stackrel{\textit{what is \underline{y} when x = 5? let's use the 1st equation}}{y=(5)^2+2(5)\implies y=25+10\implies 35}\qquad \qquad\qquad (5,35) \\\\\\ \stackrel{\textit{what is \underline{y} when x = -4? let's use the 2nd equation}}{y=3(-4)+20\implies y=-12+20\implies y=8}\qquad \qquad (-4,8)`