# How do you solve this system of equations: y= \sqrt { 2x + 1} and y = \sqrt { 3- 4x }?

Jan 12, 2018

$x = \frac{1}{3}$ and $y = \sqrt{\frac{5}{3}}$

#### Explanation:

We have that:

$y = \sqrt{2 x + 1}$ and $y = \sqrt{3 - 4 x}$

Begin by setting $y = y$ so that:

$\sqrt{2 x + 1} = \sqrt{3 - 4 x}$

Now square:

$2 x + 1 = 3 - 4 x$

$\to 6 x = 2 \therefore x = \frac{1}{3}$

Now $y$; substitute into either equation for $y$:

$y = \sqrt{2 x + 1} = \sqrt{2 \left(\frac{1}{3}\right) + 1} = \sqrt{\frac{2}{3} + 1} = \sqrt{\frac{5}{3}}$

We can check if we are correct with the other equation:

$y = \sqrt{3 - 4 \left(\frac{1}{3}\right)} = \sqrt{3 - \frac{4}{3}} = \sqrt{\frac{5}{3}}$

So $x = \frac{1}{3}$ and $y = \sqrt{\frac{5}{3}}$