# How do you solve 4x + 5y = 1 and 7y - 3x = 2?

Sep 21, 2015

$x = - \frac{3}{43}$
$y = \frac{11}{43}$

#### Explanation:

There are two ways you can solve this: substitution or elimination. I'll go with substitution since I personally like it better.

Let's try looking for the value of $x$ first. We'll first try to isolate $y$ in the first equation.
$4 x + 5 y = 1$
$5 y = 1 - 4 x$
$y = \frac{1 - 4 x}{5}$

Now we will substitute this value of $y$ to the second equation.
$7 y - 3 x = 2$
$7 \left(\frac{1 - 4 x}{5}\right) - 3 x = 2$
(We'll multiply both sides by 5 to remove the denominator)
$\left(5\right) \left[7 \left(\frac{1 - 4 x}{5}\right) - 3 x\right] = \left(5\right) \left(2\right)$
$7 \left(1 - 4 x\right) - 15 x = 10$
$7 - 28 x - 15 x = 10$
$7 - 43 x = 10$
$- 43 x = 10 - 7$
$- 43 x = 3$
$x = - \frac{3}{43}$

To solve for $y$, substitute the value of $x$ we just got into either of the two equations.
$4 x + 5 y = 1$
$4 \left(- \frac{3}{43}\right) + 5 y = 1$
$- \frac{12}{43} + 5 y = 1$
$5 y = 1 + \frac{12}{43}$
$5 y = \frac{43}{43} + \frac{12}{43}$
$5 y = \frac{55}{43}$
$\left(\frac{1}{5}\right) 5 y = \left(\frac{1}{5}\right) \frac{55}{43}$
$y = \frac{11}{43}$