# How do you solve the following system?:  2x - 4y = 5 , x-4y=3

Nov 13, 2015

I found:
$x = 2$
$y = - \frac{1}{4}$

#### Explanation:

I would isolate $x$ from the second equation and substitute it into the first:
$x = 3 + 4 y$
and so:
$2 \left(\textcolor{red}{3 + 4 y}\right) - 4 y = 5$ solve for $y$:
$6 + 8 y - 4 y = 5$
$4 y = - 1$
$y = - \frac{1}{4}$
Substitute this back into: $x = 3 + 4 y$
$x = 3 + 4 \left(\textcolor{red}{- \frac{1}{4}}\right) = 3 - 1 = 2$

Nov 13, 2015

Use substitution for $x$ then plug back in with $y$.

#### Explanation:

If $x - 4 y = 3$, then add $4 y$ to both sides to get $x = 4 y + 3$.

Knowing this, you can plug in $4 y + 3$ in the other equation in place of $x$ since they are equivalent.

This results in: $2 \left(4 y + 3\right) - 4 y = 5$
Distribute the 2 to get: $8 y + 6 - 4 y = 5$
Combine like terms on the left: $4 y + 6 = 5$
Subtract 6 from both sides: $4 y = - 1$
Divide both sides by 4: $\textcolor{red}{y = - \frac{1}{4}}$

With this knowledge, plug $y$ into either equation to determine the value of $x$, as such: $x - 4 \left(- \frac{1}{4}\right) = 3$
Multiply (remember that a negative time a negative is positive): $x + 1 = 3$
Add 1 to both sides: $\textcolor{red}{x = 2}$

Thus, our ordered pair is $\left(2 , - \frac{1}{4}\right)$.