Step-by-step explanation
We need to solve the following system of equations:
![\begin{gathered} y=-8x \\ 2x+4y=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/tj0y5ml8712h6pfjwm2l6otb3l69tmyh9y.png)
We can use the substitution method. The first equation gives us an expression of y dependent on x. We can take the second equation and replace y by that expression so we obtain an equation with only one variable:
![\begin{gathered} 2x+4y=0 \\ 2x+4\cdot(-8x)=0 \\ 2x-32x=0 \\ (2-32)x=0 \\ -30x=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/4p264mjxj7ye8g74i7cvam384z8lijw0pc.png)
Then we divide both sides by -30:
![\begin{gathered} (-30x)/(-30)=(0)/(-30) \\ x=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/pw79lj85lsmawtvwivmuef4w8v2oup2shj.png)
So we have found that x=0. We use this value in the expression of y:
![y=-8x=-8\cdot0=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/sap4ygvg546ub7g3aoc32hxfb5tn1e4ked.png)
So we have x=0 and y=0.
Answer
Then the answer is (0,0).