Given the system of equations:
[tex]\begin{gathered} x^2+y^2=68 \\ y=4x \end{gathered}[/tex]
We can solve it by substituting the second equation into the first as follows:
[tex]\begin{gathered} x^2+(4x)^2=68 \\ \\ Operating: \\ \\ x^2+16x^2=68 \\ 17x^2=68 \end{gathered}[/tex]
Dividing by 17:
[tex]x^2=\frac{68}{17}=4[/tex]
Applying square root on both sides:
[tex]\begin{gathered} x=\pm\sqrt{4} \\ x=\pm2 \end{gathered}[/tex]
There are two solutions for x and they produce two solutions for y.
For x = 2
y = 4*2 = 8
For x = -2
y = 4*(-2) = -8
Thus, the solutions are:
(2,8) and (-2, -8)
