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Question

Solve each system by the substitution method . Be sure to check all proposed solutions.

{x+8y=62x+4y=3\left\{ \begin{array} { r } { x + 8 y = 6 } \\ { 2 x + 4 y = - 3 } \end{array} \right.

Solution

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We are given:

{x+8y=62x+4y=3\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 xx having a coefficient of 1 so we can isolate it to rewrite it as:

x=68yx=6-8y

Substitute to the second equation and solve for yy:

2(68y)+4y=32(6-8y)+4y=-3

1216y+4y=312-16y+4y=-3

12y=15-12y=-15

y=54\color{#4257b2}y=\dfrac{5}{4}

Solve for xx using the rewritten first equation:

x=68(54)x=6-8\left(\dfrac{5}{4}\right)

x=610x=6-10

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

Check:

4+8(54)=?62(4)+4(54)=?34+10=?68+5=?36=63=3\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:

{(4,54)}(1)\color{#c34632}\left\{\left (-4,\dfrac{5}{4}\right) \right \}\color{white}\tag{1}

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