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Question

# Solve each system by the substitution method . Be sure to check all proposed solutions.$\left\{ \begin{array} { r } { x + 8 y = 6 } \\ { 2 x + 4 y = - 3 } \end{array} \right.$

Solution

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Step 1
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We are given:

$\begin{cases} x+8y=6\\ 2x+4y=-3\\ \end{cases}$

No variable is isolated in either of the two equations but the first equation has $x$ having a coefficient of 1 so we can isolate it to rewrite it as:

$x=6-8y$

Substitute to the second equation and solve for $y$:

$2(6-8y)+4y=-3$

$12-16y+4y=-3$

$-12y=-15$

$\color{#4257b2}y=\dfrac{5}{4}$

Solve for $x$ using the rewritten first equation:

$x=6-8\left(\dfrac{5}{4}\right)$

$x=6-10$

$\color{#4257b2}x=-4$

Check:

\begin{align*} -4+8\left(\dfrac{5}{4}\right) &\stackrel{?}{=}6 & 2(-4)+4\left(\dfrac{5}{4}\right) &\stackrel{?}{=}-3\\ -4+10 &\stackrel{?}{=}6 & -8+5&\stackrel{?}{=}-3\\ 6&=6\hspace{5mm}\checkmark & -3&=-3\hspace{5mm}\checkmark \end{align*}

Both equations of the system are satisfied so the solution set of the system is:

$\color{#c34632}\left\{\left (-4,\dfrac{5}{4}\right) \right \}\color{white}\tag{1}$

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