Mathematics, 12.07.2019 17:30 jadabugs8187
How do you solve systems of equations but substitution? .
Answers: 1
![\bf \begin{cases} -7x-2y=13\\[-0.5em] \hrulefill\\ x-2y=11\\ \boxed{x}=11+2y \end{cases}~\hspace{5em}\stackrel{\textit{substitution \boxed{x} on the second equation}}{-7\left( \boxed{11+2y} \right)-2y=13} \\\\\\ -77-14y-2y=13\implies -77-16y=13\implies -90=16y \\\\\\ \cfrac{-90}{16}=y\implies \blacktriangleright -\cfrac{45}{8}=y \blacktriangleleft](/tpl/images/0081/7224/5d8e4.png)
![\bf ~\dotfill\\\\ \stackrel{\textit{substituting that \underline{y} in the second equation}}{x-2\left( \boxed{-\frac{45}{8}} \right)=11}\implies x+\cfrac{45}{4}=11 \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4x+45=44}\implies 4x=-1\implies \blacktriangleright x=-\cfrac{1}{4} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hspace{15em}\left(-\frac{1}{4}~,~-\frac{45}{8} \right)](/tpl/images/0081/7224/3e81d.png)
![\bf \begin{cases} y=7x-10\\ y=-3 \end{cases}~\hspace{1em}\stackrel{\textit{since we know y = -3, we plugged that in the first equation}}{-3=7x-10\implies 7=7x\implies \cfrac{7}{7}=x\implies \blacktriangleright 1=x \blacktriangleleft}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hspace{15em}(1,-3)](/tpl/images/0081/7224/cb617.png)
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How do you solve systems of equations but substitution? .[tex] - 7x - 2y = 13 \\ x - 2y = 11[/tex]...
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