# Determine whether the equation defines y as a function of x. `7x^2+3y^2=1`

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First we will have to isolate y on the left side.

`==gt 7x^2 + 3y^2 = 1`

` ==gt 3y^2 = 1- 7x^2 `

`==gt y^2 = (1-7x^2)/3 `

`==gt y= +- (1-7x^2)/3`

`` Now we notice that we have two possible values to y for each value of x.

Therefore, we have two images for the value of x ( positive value and negative value.

For example, if x= 1 ==> y= 2 and y= -2)

Then, y is **NOT** a function of x.

Or, we can graph y and test the vertical line.

Isolate y so we can find out if this is indeed a function or not:

`3y^2 = 1 - 7x^2`

`y^2 = ( 1 - 7x^2 ) / 3`

`y = sqrt ( (1 - 7x^2) / 3 )`

` `

We know that the square root means both positive and negative. So when we plug this into a graphing utility, we need to account for both positive and negative. And when you do this, you'll see that the resulting image is that of a circle which does not pass the vertical line test and so is not a function.