[tex]\bf \begin{cases} x +y = 4\\ 7x + 9y = 42 \end{cases}\qquad \qquad \stackrel{\textit{solving for "y" on the 1st equation}}{\boxed{y} = 4-x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{7x+9\left(\boxed{4-x} \right) = 42}\implies 7x+36-9x=42\implies -2x+36=42 \\\\\\ -2x=6\implies x = \cfrac{6}{-2}\implies \blacktriangleright x = -3 \blacktriangleleft[/tex]
[tex]\bf \stackrel{\textit{since we know that}}{y = 4-x}\implies y = 4-(-3)\implies y = 4+3\implies \blacktriangleright y = 7 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (-3~~,~~7)~\hfill[/tex]
