Question

y = x2 + 5x - 10y=-x² + 2x + 10

Answer 1

Since in both given equations, the variable y is already clear, then you can equal the two equations and then solve for x. So, you have

$\begin{cases}y=x^2+5x-10\text{ (1)} \\ y=-x^{2}+2x+10\text{ (2)}\end{cases}$$\begin{gathered} x^2+5x-10=-x^2+2x+10 \\ \text{ Add }x^2\text{ to both sides of the equation} \\ \text{ Subtract 2x to both sides of the equation} \\ \text{ Subtract 10 to both sides of the equation} \\ x^2+5x-10+x^2-2x-10=-x^2+2x+10+x^2-2x-10 \\ 2x^2+3x-20=0 \end{gathered}$

To solve for x you can use the quadratic formula, that is,

$\begin{gathered} \text{ For }ax^2+bx+c=0\text{ where a}\ne0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

In this case

a=2

b=3

c=-20

So,

$\begin{gathered} x=\frac{-3\pm\sqrt[]{(3)^2-4(2)(-20)}}{2\cdot2} \\ x=\frac{-3\pm\sqrt[]{9+160}}{4} \\ x=\frac{-3\pm\sqrt[]{169}}{4} \\ x=\frac{-3\pm13}{4} \\ x_1=\frac{-3+13}{4}=\frac{10}{4}=\frac{5}{2}=2.5 \\ x_2=\frac{-3-13}{4}=\frac{-16}{4}=-4 \end{gathered}$

Now you can plug in the solutions found in any of the given equations to find their respective y-coordinates.

For the first solution, you have

$\begin{gathered} x_1=\frac{5}{2} \\ y_{}=-x^2+2x+10\text{ (2)} \\ y_1=-(\frac{5}{2})^2+2(\frac{5}{2})+10 \\ y_1=-\frac{25}{4}+5+10 \\ y_1=\frac{35}{4} \\ y_1=8.75 \\ \text{ Then} \\ (2.5,8.75) \end{gathered}$

For the second solution, you have

$\begin{gathered} x_2=-4 \\ y=-x^2+2x+10\text{ (2)} \\ y_2=-(-4)^2+2(-4)+10 \\ y_2=-16-8+10 \\ y_2=-14 \\ \text{Then} \\ (-4,-14) \end{gathered}$

Therefore, the solution set of the given system of equations is

$\mleft\lbrace(-4,-14),(2.5,8.75)\mright\rbrace$