Question

1. \\( 5x + 4y = -20 \\)

\\( y = -5 \\)
\\( (\quad, \quad) \\) sum of x & y \\( \quad \\) building \\( \quad \\)

1. \\( -7x + 4y = -20 \\)

\\( 4x - 4y = 0 \\)
\\( (\quad, \quad) \\) sum of x & y \\( \quad \\) building \\( \quad \\)

Explanation:

Problem 9:

Step1: Substitute \( y = -5 \) into the first equation

We have the equation \( 5x + 4y = -20 \) and \( y = -5 \). Substitute \( y \) into the first equation:
\( 5x + 4(-5) = -20 \)

Step2: Simplify and solve for \( x \)

Simplify the left - hand side: \( 5x-20=-20 \)
Add 20 to both sides: \( 5x-20 + 20=-20 + 20 \), which gives \( 5x = 0 \)
Divide both sides by 5: \( x=\frac{0}{5}=0 \)

Step3: Find the sum of \( x \) and \( y \)

We know that \( x = 0 \) and \( y=-5 \), so the sum \( x + y=0+(-5)=-5 \)

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

We have the system of equations:
\(

$$\begin{cases}-7x + 4y=-20\\4x-4y = 0\end{cases}$$

\)
Add the two equations together: \( (-7x + 4y)+(4x - 4y)=-20 + 0 \)

Step2: Simplify and solve for \( x \)

Simplify the left - hand side: \( -7x+4y + 4x-4y=-3x \)
Simplify the right - hand side: \( -20 \)
So, \( -3x=-20 \), divide both sides by - 3: \( x=\frac{-20}{-3}=\frac{20}{3} \)

Step3: Substitute \( x=\frac{20}{3} \) into one of the equations to solve for \( y \)

We use the equation \( 4x-4y = 0 \). Substitute \( x=\frac{20}{3} \):
\( 4\times\frac{20}{3}-4y = 0 \)
\( \frac{80}{3}-4y = 0 \)
Add \( 4y \) to both sides: \( \frac{80}{3}=4y \)
Divide both sides by 4: \( y=\frac{80}{3}\div4=\frac{80}{3}\times\frac{1}{4}=\frac{20}{3} \)

Step4: Find the sum of \( x \) and \( y \)

We know that \( x=\frac{20}{3} \) and \( y = \frac{20}{3} \), so the sum \( x + y=\frac{20}{3}+\frac{20}{3}=\frac{40}{3} \)

Answer:

The ordered pair is \( (0,-5) \), the sum of \( x \) and \( y \) is \( -5 \). (The "Building" part is not clear from the given problem, so we can't provide information for it.)

Problem 10: