what are the solutions of the following system?\n$\begin{cases}x^{2}+y^{2}=25\\2x + y=-5end{cases}$\n(0, -5)…

Explanation:

Step1: Isolate y in the second - equation

From $2x + y=-5$, we get $y=-2x - 5$.

Step2: Substitute y into the first - equation

Substitute $y=-2x - 5$ into $x^{2}+y^{2}=25$. Then $x^{2}+(-2x - 5)^{2}=25$. Expand $(-2x - 5)^{2}$ using $(a + b)^{2}=a^{2}+2ab + b^{2}$, where $a=-2x$ and $b = - 5$. So $(-2x - 5)^{2}=4x^{2}+20x + 25$. The equation becomes $x^{2}+4x^{2}+20x + 25 = 25$. Combine like - terms: $5x^{2}+20x+25 - 25 = 0$, which simplifies to $5x^{2}+20x=0$. Factor out $5x$: $5x(x + 4)=0$.

Step3: Solve for x

Set each factor equal to zero. If $5x=0$, then $x = 0$. If $x + 4=0$, then $x=-4$.

Step4: Solve for y

When $x = 0$, substitute into $y=-2x - 5$, we get $y=-2\times0 - 5=-5$. When $x=-4$, substitute into $y=-2x - 5$, we get $y=-2\times(-4)-5=8 - 5 = 3$.

Answer:

$(0, - 5)$ and $(-4,3)$