Question

elimination
1
6x-7y=22
-4x+3y= -8

2
28x-28y= -28
16x-16y=-16

3

-2x+6y=10
-5x+5y=-25

Answer 1

Let’s solve the systems of equations you provided one by one using the elimination method.

### System 1:
1. \( 6x - 7y = 22 \) (Equation 1)
2. \( -4x + 3y = -8 \) (Equation 2)

To eliminate \(y\), we can multiply Equation 1 by 3 and Equation 2 by 7, so that the coefficients of \(y\) will match.

\[
3(6x - 7y) = 3(22) \implies 18x - 21y = 66 \quad (Equation 3)
\]
\[
7(-4x + 3y) = 7(-8) \implies -28x + 21y = -56 \quad (Equation 4)
\]

Now we can add Equation 3 and Equation 4 to eliminate \(y\):

\[
(18x - 21y) + (-28x + 21y) = 66 - 56
\]
\[
-10x = 10 \implies x = -1
\]

Now substitute \(x = -1\) back into one of the original equations to solve for \(y\). Using Equation 1:

\[
6(-1) - 7y = 22
\]
\[
-6 - 7y = 22 \implies -7y = 22 + 6 \implies -7y = 28
\]
\[
y = -4
\]

**Solution to System 1: \( (x, y) = (-1, -4) \)**

### System 2:
1. \( 28x - 28y = -28 \) (Equation 1)
2. \( 16x - 16y = -16 \) (Equation 2)

We can simplify both equations first:

\[
28x - 28y = -28 \implies x - y = -1 \quad (Equation 3)
\]
\[
16x - 16y = -16 \implies x - y = -1 \quad (Equation 4)
\]

Both simplified equations are the same:

\[
x - y = -1
\]

This means there are infinitely many solutions where \(y = x + 1\). The solution can be expressed as:

**Solution to System 2: \( (x, y) = (t, t + 1) \) for any real number \(t\)**

### System 3:
1. \( -2x + 6y = 10 \) (Equation 1)
2. \( -5x + 5y = -25 \) (Equation 2)

To eliminate \(y\), we can make the coefficients of \(y\) equal. We can multiply Equation 1 by 5 and Equation 2 by 6:

\[
5(-2x + 6y) = 5(10) \implies -10x + 30y = 50 \quad (Equation 3)
\]
\[
6(-5x + 5y) = 6(-25) \implies -30x + 30y = -150 \quad (Equation 4)
\]

Now we can subtract Equation 3 from Equation 4:

\[
(-30x + 30y) - (-10x + 30y) = -150 - 50
\]
\[
-20x = -200 \implies x = 10
\]

Now substitute \(x = 10\) back into one of the original equations to solve for \(y\). Using Equation 1:

\[
-2(10) + 6y = 10
\]
\[
-20 + 6y = 10 \implies 6y = 30 \implies y = 5
\]

**Solution to System 3: \( (x, y) = (10, 5) \)**

### Summary of Solutions:
1. **System 1**: \( (x, y) = (-1, -4) \)
2. **System 2**: \( (x, y) = (t, t + 1) \) for any real number \(t\)
3. **System 3**: \( (x, y) = (10, 5) \)