Question

Resolva os sistemas de equações abaixo utilizando o método da adição:

Answer 1

$\boxed{\begin{array}{l}\tt c)~\begin{cases}\sf 4x-y=-4\\\sf 5x-y=1\end{cases}\\\underline{\rm multiplique~a~1^a~equac_{\!\!,}\tilde ao~por~-1}\\\begin{cases}\sf4x-y=-4\cdot(-1)\\\sf5x-y=1\end{cases}\\\underline{\rm somando~algebricamente~temos:}\\+\underline{\begin{cases}\sf-4x+\backslash\!\!\!y=4\\\sf5x-\backslash\!\!\!y=1\end{cases}}\\\sf x=5\\\underline{\rm substituindo~na~2^a~equac_{\!\!,}\tilde ao~temos:}\\\sf 5\cdot5-y=1\\\sf 25-y=1\\\sf -y=1-25\\\sf -y=-24\cdot(-1)\\\sf y=24\end{array}}$

$\Large\boxed{\begin{array}{l}\sf S=\{(5,24)\}\end{array}}$

$\boxed{\begin{array}{l}\tt d)~\begin{cases}\sf x+2y=6\\\sf x-y=2\end{cases}\\\underline{\rm multiplique~por~2~a~2^a~equac_{\!\!,}\tilde ao:}\\\begin{cases}\sf x+2y=6\\\sf2x-2y=4\end{cases}\\\underline{\rm adicione~as~equac_{\!\!,}\tilde oes~algebricamente:}\\+\underline{\begin{cases}\sf x+\diagdown\!\!\!\!\!2y=6\\\sf2x-\diagdown\!\!\!\!\!\!2y=4\end{cases}}\\\sf 3x=10\\\sf x=\dfrac{10}{3}\\\underline{\rm substituindo~na~2^a~equac_{\!\!,}\tilde ao~temos:}\\\sf \dfrac{10}{3}-y=2\cdot3\\\sf 10-3y=6\end{array}}$

$\large\boxed{\begin{array}{l}\sf 3y=10-6\\\sf 3y=4\\\sf y=\dfrac{4}{3}\\\\\sf S=\bigg\{\bigg(\dfrac{10}{3},\dfrac{4}{3}\bigg)\bigg\}\end{array}}$