Question

In the system shown below, what are the coordinates of the solution that lies in quadrant IV?

Write your answer in the form (a,b) without using spaces


`2x^2+y^2=33\\x^2+y^2+2y=19`

Answer 1

The coordinates of the solution that lies in quadrant IV are (2, -5)

Explanation:

We have 2 equations, the first of an ellipse and the second of a circumference.

`2x^2+y^2=33\\x^2+y^2+2y=19`

To solve the system solve the second equation for x and then substitute in the first equation

`x^2+y^2+2y=19\\\\x^2 = 19 -y^2 -2y`

So Substituting in the first equation we have

`x^2 = 19 -y^2 -2y\\\\2(19 -y^2 -2y)+y^2=33\\\\38 -2y^2-4y +y^2 = 33\\\\-y^2-4y+5=0\\\\y^2 +4y-5 = 0`

Now we must factor the quadratic expression.

We look for two numbers that multiply as a result -5 and add them as result 4.

These numbers are -1 and 5.

Then the factors are

`y^2 +4y-5 = 0\\\\(y-1)(y+5) = 0`

Therefore the system solutions are:

`y = 1`;
`y = -5`

In the 4th quadrant the values of x are positive and the values of y are negative.

So we take the negative value of y and substitute it into the system equation to find x

`y=-5\\\\2x^2+(-5)^2=33\\\\2x^2 = 33-25\\\\2x^2 = 8\\\\x^2 = 4`

`x = 2`, and
`x= -2`

In the 4th quadrant the values of x are positive

So we take the positive value of x

the coordinates of the solution that lies in quadrant IV are (2, -5)