Question

In the system shown below what is the sum of the X coordinates of all solutions?

Answer 1

Given the system of equations:

`\begin{gathered} x^2+4y^2=100 \\ \\ 4y-x^2=-20 \end{gathered}`

Let's find the sum of the x-coordinates of all solutions.

Let's solve the system simultaneously using substitution method to find the solution.

• Rewrite the second equation for x²:

`x^2=4y+20`

• Substitute 4y + 20 for x² in the first equation:

`4y+20+4y^2=100`

Now, let's find the values of y

Subtract 20 from both sides

`\begin{gathered} 4y+20-20+4y^2=100-20 \\ \\ 4y+4y^2=80 \\ \\ 4y+4y^2-80=0 \\ \\ Factor\text{ out 4:} \\ 4(y+y^2-20)=0 \\ Factor\text{ using the AC method:} \\ 4(y-4)(y+5)=0 \end{gathered}`

Set each factor to zero and solve for y:

`\begin{gathered} y-4=0 \\ Add\text{ 4 to both sides:} \\ y-4+4=0+4 \\ y=4 \\ \\ \\ y+5=0 \\ Subtract\text{ 5 from both sides:} \\ y+5-5=0-5 \\ y=-5 \end{gathered}`

We have the solutions for y:

y = 4 and -5

Now, let's find the values of x when y = 4 and -5

• When y = 4:

Substitute 4 for y in either of the equations and solve for x.

Take equation 1:

`\begin{gathered} x^2+4y^2=100 \\ x^2+4(4)^2=100 \\ x^2+4(16)=100 \\ x^2+64=100 \\ x^2=100-64 \\ x^2=36 \\ Take\text{ the square root of both sides:} \\ √(x^2)=\pm√(36) \\ \\ x=-6,\text{ 6} \end{gathered}`

• When y = -5:

`\begin{gathered} x^2+4(-5)^2=100 \\ x^2+4(25)=100 \\ x^2+100=100 \\ x^2=100-100 \\ x=0 \end{gathered}`

Therefore, the values of x are:

-6, 6, 0

The sum of the x-coordinates is:

-6 + 6 + 0 = 0

ANSWER:

0