Question

1. write the equation in the standard form of a circle $(x - h)^2+(y - k)^2 = r^2$,$x^2 + y^2+2x - 10y + 10 = 0$,a.) $(x + 1)^2+(y - 5)^2 = 16$ c.) $(x - 1)^2+(y + 5)^2 = 16$b.) $(x + 1)^2+(y - 5)^2 = 4$ d.) $(x - 1)^2+(y + 5)^2 = 4$43. what is the solution to the system graphed?a.) $(2,-1)$ b.) $(-1,2)$ c.) $(1,-2)$ d.) $(-2,1)$44. what is the solution to the system of equations?$\begin{cases}y = 2x - 1\\y=-x + 5end{cases}$a.) $(2,3)$ b.) $(-3,2)$ c.) $(3,2)$ d.) $(-2,-3)$

Explanation:

Step1: Solve the system of equations

Given the system \(

$$\begin{cases}y = 2x-1\\y=-x + 5\end{cases}$$

\), since both expressions equal \(y\), we can set them equal to each other:
\(2x-1=-x + 5\)

Step2: Solve for \(x\)

Add \(x\) to both sides: \(2x+x-1=-x+x + 5\), which simplifies to \(3x-1=5\).
Add \(1\) to both sides: \(3x-1 + 1=5+1\), so \(3x=6\).
Divide both sides by \(3\): \(x=\frac{6}{3}=2\).

Step3: Solve for \(y\)

Substitute \(x = 2\) into \(y = 2x-1\): \(y=2\times2-1=4 - 1=3\).

Answer:

A. \((2,3)\)