# The square of x is equal to 4 times the square of y. If x is 1 more than twice y, what is the value of x?

Feb 9, 2016

$x = \frac{1}{2}$, $y = - \frac{1}{4}$

#### Explanation:

Let's describe the situation in equations.

The first sentence can be written as

${x}^{2} = 4 {y}^{2}$

and the second one as

$x = 1 + 2 y$

So now we have two equations that we can solve for $x$ and $y$.

To do so, let's plug the second equation into the first equation, so plug $1 + 2 y$ for every occurence of $x$ in the first equation:

${\left(1 + 2 y\right)}^{2} = 4 {y}^{2}$

$1 + 4 y + 4 {y}^{2} = 4 {y}^{2}$

... subtract $4 {y}^{2}$ on both sides...

$1 + 4 y = 0$

... subtract $1$ on both sides...

$4 y = - 1$

...divide by $4$ on both sides...

$y = - \frac{1}{4}$

Now that we have $y$, we can plug the value into the second equation to find $x$:

$x = 1 + 2 \cdot \left(- \frac{1}{4}\right) = 1 - \frac{1}{2} = \frac{1}{2}$

===================

You can make a quick check if $x$ and $y$ were computed correctly:

• the square of $x$ is ${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$, the square of $y$ is ${\left(- \frac{1}{4}\right)}^{2} = \frac{1}{16}$. The square of $x$ is indeed equal to $4$ times the square of $y$.
• twice $y$ is $- \frac{1}{2}$, and one more is $- \frac{1}{2} + 1 = \frac{1}{2}$ which is indeed $x$.