Question

given each system of linear equations below, solve for the solution. type your answer in the space provided. note: do not add extra pacing (x,y) 1. 2x - 4y = -2 -8x + 4y = 8 2. -2x + 2y = -14 2x + 6y = -10 3. 9x - 4y = -20 -3x + 4y = 20 4. -5x + 7y = 12 -5x + 6y = 11

Explanation:

Problem 1

Step1: Add the two equations to eliminate \( y \)

The system is:
\[

$$\begin{cases} 2x - 4y = -2 \\ -8x + 4y = 8 \end{cases}$$

\]
Adding the two equations: \((2x - 4y)+(-8x + 4y)=-2 + 8\)
Simplify: \(2x-8x-4y + 4y=6\) which gives \(-6x=6\)

Step2: Solve for \( x \)

Divide both sides by \(-6\): \(x=\frac{6}{-6}=-1\)

Step3: Substitute \( x = -1 \) into the first equation to find \( y \)

Substitute into \(2x - 4y=-2\): \(2(-1)-4y=-2\)
Simplify: \(-2-4y=-2\)
Add 2 to both sides: \(-4y=0\)
Divide by \(-4\): \(y = 0\)

Step1: Add the two equations to eliminate \( x \)

The system is:
\[

$$\begin{cases} -2x + 2y = -14 \\ 2x + 6y = -10 \end{cases}$$

\]
Adding the two equations: \((-2x + 2y)+(2x + 6y)=-14-10\)
Simplify: \(-2x+2x+2y + 6y=-24\) which gives \(8y=-24\)

Step2: Solve for \( y \)

Divide both sides by 8: \(y=\frac{-24}{8}=-3\)

Step3: Substitute \( y = -3 \) into the second equation to find \( x \)

Substitute into \(2x + 6y=-10\): \(2x+6(-3)=-10\)
Simplify: \(2x-18=-10\)
Add 18 to both sides: \(2x = 8\)
Divide by 2: \(x = 4\)

Step1: Add the two equations to eliminate \( y \)

The system is:
\[

$$\begin{cases} 9x - 4y = -20 \\ -3x + 4y = 20 \end{cases}$$

\]
Adding the two equations: \((9x - 4y)+(-3x + 4y)=-20 + 20\)
Simplify: \(9x-3x-4y + 4y=0\) which gives \(6x=0\)

Step2: Solve for \( x \)

Divide both sides by 6: \(x = 0\)

Step3: Substitute \( x = 0 \) into the first equation to find \( y \)

Substitute into \(9x - 4y=-20\): \(9(0)-4y=-20\)
Simplify: \(-4y=-20\)
Divide by \(-4\): \(y = 5\)

Answer:

\((-1,0)\)

Problem 2