Question

1. $-4x + 10y = -4$\\

$(-10x + y = -10)(-10)$\\
\\

1. $4x + 10y = -20$\\

$-3x - 5y = 5$

Explanation:

Problem 11: Solve the system of equations

We have the system:
\[

$$\begin{cases} -4x + 10y = -4 \\ -10x + y = -10 \end{cases}$$

\]
And a note to multiply the second equation by \(-10\). Let's follow that.

Step 1: Multiply the second equation by \(-10\)

The second equation is \(-10x + y = -10\). Multiply each term by \(-10\):
\[
(-10x)\times(-10) + y\times(-10) = (-10)\times(-10)
\]
\[
100x - 10y = 100
\]

Step 2: Add the first equation to the new second equation

The first equation is \(-4x + 10y = -4\). Add it to \(100x - 10y = 100\):
\[
(-4x + 10y) + (100x - 10y) = -4 + 100
\]
Simplify the left side: \(-4x + 100x + 10y - 10y = 96x\)
Simplify the right side: \(96\)
So we have \(96x = 96\)

Step 3: Solve for \(x\)

Divide both sides by \(96\):
\[
x = \frac{96}{96} = 1
\]

Step 4: Substitute \(x = 1\) into one of the original equations to find \(y\)

Let's use the second original equation: \(-10x + y = -10\)
Substitute \(x = 1\):
\[
-10(1) + y = -10
\]
\[
-10 + y = -10
\]
Add \(10\) to both sides:
\[
y = -10 + 10 = 0
\]

Problem 13: Solve the system of equations

We have the system:
\[

$$\begin{cases} 4x + 10y = -20 \\ -3x - 5y = 5 \end{cases}$$

\]

Step 1: Multiply the second equation by \(2\) to eliminate \(y\)

The second equation is \(-3x - 5y = 5\). Multiply each term by \(2\):
\[
-6x - 10y = 10
\]

Step 2: Add the first equation to the new second equation

The first equation is \(4x + 10y = -20\). Add it to \(-6x - 10y = 10\):
\[
(4x + 10y) + (-6x - 10y) = -20 + 10
\]
Simplify the left side: \(4x - 6x + 10y - 10y = -2x\)
Simplify the right side: \(-10\)
So we have \(-2x = -10\)

Step 3: Solve for \(x\)

Divide both sides by \(-2\):
\[
x = \frac{-10}{-2} = 5
\]

Step 4: Substitute \(x = 5\) into one of the original equations to find \(y\)

Let's use the first original equation: \(4x + 10y = -20\)
Substitute \(x = 5\):
\[
4(5) + 10y = -20
\]
\[
20 + 10y = -20
\]
Subtract \(20\) from both sides:
\[
10y = -20 - 20 = -40
\]
Divide both sides by \(10\):
\[
y = \frac{-40}{10} = -4
\]

Final Answers

For problem 11: \(x = 1\), \(y = 0\) (or the solution is \((1, 0)\))
For problem 13: \(x = 5\), \(y = -4\) (or the solution is \((5, -4)\))

Answer:

Problem 11: Solve the system of equations

We have the system:
\[

$$\begin{cases} -4x + 10y = -4 \\ -10x + y = -10 \end{cases}$$

\]
And a note to multiply the second equation by \(-10\). Let's follow that.

Step 1: Multiply the second equation by \(-10\)

The second equation is \(-10x + y = -10\). Multiply each term by \(-10\):
\[
(-10x)\times(-10) + y\times(-10) = (-10)\times(-10)
\]
\[
100x - 10y = 100
\]

Step 2: Add the first equation to the new second equation

The first equation is \(-4x + 10y = -4\). Add it to \(100x - 10y = 100\):
\[
(-4x + 10y) + (100x - 10y) = -4 + 100
\]
Simplify the left side: \(-4x + 100x + 10y - 10y = 96x\)
Simplify the right side: \(96\)
So we have \(96x = 96\)

Step 3: Solve for \(x\)

Divide both sides by \(96\):
\[
x = \frac{96}{96} = 1
\]

Step 4: Substitute \(x = 1\) into one of the original equations to find \(y\)

Let's use the second original equation: \(-10x + y = -10\)
Substitute \(x = 1\):
\[
-10(1) + y = -10
\]
\[
-10 + y = -10
\]
Add \(10\) to both sides:
\[
y = -10 + 10 = 0
\]

Problem 13: Solve the system of equations

We have the system:
\[

$$\begin{cases} 4x + 10y = -20 \\ -3x - 5y = 5 \end{cases}$$

\]

Step 1: Multiply the second equation by \(2\) to eliminate \(y\)

The second equation is \(-3x - 5y = 5\). Multiply each term by \(2\):
\[
-6x - 10y = 10
\]

Step 2: Add the first equation to the new second equation

The first equation is \(4x + 10y = -20\). Add it to \(-6x - 10y = 10\):
\[
(4x + 10y) + (-6x - 10y) = -20 + 10
\]
Simplify the left side: \(4x - 6x + 10y - 10y = -2x\)
Simplify the right side: \(-10\)
So we have \(-2x = -10\)

Step 3: Solve for \(x\)

Divide both sides by \(-2\):
\[
x = \frac{-10}{-2} = 5
\]

Step 4: Substitute \(x = 5\) into one of the original equations to find \(y\)

Let's use the first original equation: \(4x + 10y = -20\)
Substitute \(x = 5\):
\[
4(5) + 10y = -20
\]
\[
20 + 10y = -20
\]
Subtract \(20\) from both sides:
\[
10y = -20 - 20 = -40
\]
Divide both sides by \(10\):
\[
y = \frac{-40}{10} = -4
\]

Final Answers

For problem 11: \(x = 1\), \(y = 0\) (or the solution is \((1, 0)\))
For problem 13: \(x = 5\), \(y = -4\) (or the solution is \((5, -4)\))