# 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#?

##### 1 Answer

Feb 9, 2016

#### Explanation:

Let's describe the situation in equations.

The first sentence can be written as

#x^2 = 4y^2#

and the second one as

#x = 1 + 2y#

So now we have two equations that we can solve for

To do so, let's plug the second equation into the first equation, so plug

#(1 + 2y)^2 = 4y^2#

#1 + 4y + 4y^2 = 4y^2#

... subtract

#1 + 4y = 0#

... subtract

#4y = -1#

...divide by

#y= - 1/4#

Now that we have

#x = 1 + 2* (-1/4) = 1 - 1/2 = 1/2#

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You can make a quick check if

- the square of
#x# is#(1/2)^2 = 1/4# , the square of#y# is#(-1/4)^2 = 1/16# . The square of#x# is indeed equal to#4# times the square of#y# . - twice
#y# is#-1/2# , and one more is#-1/2 + 1 = 1/2# which is indeed#x# .