Given the system of equations:
[tex]\begin{gathered} x^2+4y^2=100 \\ \\ 4y-x^2=-20 \end{gathered}[/tex]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:
[tex]x^2=4y+20[/tex]• Substitute 4y + 20 for x² in the first equation:
[tex]4y+20+4y^2=100[/tex]Now, let's find the values of y
Subtract 20 from both sides
[tex]\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}[/tex]Set each factor to zero and solve for y:
[tex]\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}[/tex]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:
[tex]\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:} \\ \sqrt{x^2}=\pm\sqrt{36} \\ \\ x=-6,\text{ 6} \end{gathered}[/tex]• When y = -5:
[tex]\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}[/tex]Therefore, the values of x are:
-6, 6, 0
The sum of the x-coordinates is:
-6 + 6 + 0 = 0
ANSWER:
0
