How do you solve the following system: 2x-4y=6 , y + 4x = 16 ?

1 Answer
May 17, 2018

See a solution process below:

Explanation:

Step 1) Solve the second equation for $y$:

$y + 4 x = 16$

$y + 4 x - \textcolor{red}{4 x} = 16 - \textcolor{red}{4 x}$

$y + 0 = 16 - 4 x$

$y = 16 - 4 x$

Step 2) Substitute $\left(16 - 4 x\right)$ for $y$ in the first equation and solve for $x$:

$2 x - 4 y = 6$ becomes:

$2 x - 4 \left(16 - 4 x\right) = 6$

$2 x - \left(4 \cdot 16\right) + \left(4 \cdot 4 x\right) = 6$

$2 x - 64 + 16 x = 6$

$2 x + 16 x - 64 = 6$

$\left(2 + 16\right) x - 64 + \textcolor{red}{64} = 6 + \textcolor{red}{64}$

$18 x - 0 = 70$

$18 x = 70$

$\frac{18 x}{\textcolor{red}{18}} = \frac{70}{\textcolor{red}{18}}$

$x = \frac{35}{9}$

Step 3) Substitute $\frac{35}{9}$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$:

$y = 16 - 4 x$ becomes:

$y = 16 - \left(4 \times \frac{35}{9}\right)$

$y = \frac{144}{9} - \frac{140}{9}$

$y = \frac{144 - 140}{9}$

$y = \frac{4}{9}$

The Solution Is:

$x = \frac{35}{9}$ and $y = \frac{4}{9}$

Or

$\left(\frac{35}{9} , \frac{4}{9}\right)$