Question

A system of equations is shown

Answer 1

$\bf \begin{cases} \boxed{y}=3x-2\\ y=x^2 \end{cases}~\hspace{6em}\stackrel{\textit{doing a quick substitution}}{\boxed{3x-2}=x^2}\implies 0=x^2-3x+2 \\\\\\ 0=(x-1)(x-2)\implies \begin{cases} 0=x-1\implies &1=x\\[0.8em] 0=x-2\implies &2=x \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{using x = 1 in the first equation}}{y=3(1)-2\implies y=1}~\hfill \blacktriangleright (\stackrel{x}{1},\stackrel{y}{1}) \blacktriangleleft \\\\\\ \stackrel{\textit{using x = 2 in the first equation}}{y=3(2)-2\implies y=4}~\hfill \blacktriangleright (\stackrel{x}{2},\stackrel{y}{4})\blacktriangleleft$

Answer 2

no the first one is (2,4) and then (1,1)