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How do solve the following linear system?:  3x + 2y = 12, y = 4x + 4 ?

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Jun 19, 2018

See a solution process below:

Explanation:

Step 1) Because the second equation is already solved for $y$ we can substitute $\left(4 x + 4\right)$ for $y$ in the first equation and solve for $x$:

$3 x + 2 y = 12$ becomes:

$3 x + 2 \left(4 x + 4\right) = 12$

$3 x + \left(2 \times 4 x\right) + \left(2 \times 4\right) = 12$

$3 x + 8 x + 8 = 12$

$\left(3 + 8\right) x + 8 = 12$

$11 x + 8 = 12$

$11 x + 8 - \textcolor{red}{8} = 12 - \textcolor{red}{8}$

$11 x + 0 = 4$

$11 x = 4$

$\frac{11 x}{\textcolor{red}{11}} = \frac{4}{\textcolor{red}{11}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{11}}} x}{\cancel{\textcolor{red}{11}}} = \frac{4}{11}$

$x = \frac{4}{11}$

Step 2) Substitute $\frac{4}{11}$ for $x$ in the second equation and solve for $y$:

$y = 4 x + 4$ becomes:

$y = \left(4 \times \frac{4}{11}\right) + 4$

$y = \frac{16}{11} + 4$

$y = \frac{16}{11} + \left(4 \times \frac{11}{11}\right)$

$y = \frac{16}{11} + \frac{44}{11}$

$y = \frac{16 + 44}{11}$

$y = \frac{60}{11}$

The Solution Is:

$x = \frac{4}{11}$ and $y = \frac{60}{11}$

Or

$\left(\frac{4}{11} , \frac{60}{11}\right)$

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