Question

solve using substitution.
y = 2x + 3
3x + 2y = 12
(give details)

Answer 1

To solve the system of equations using substitution, we can start with the two equations given:

1. \( y = 2x + 3 \) (Equation 1)
2. \( 3x + 2y = 12 \) (Equation 2)

### Step 1: Substitute \(y\) in Equation 2

Since Equation 1 gives us \( y \) in terms of \( x \), we can substitute this expression for \( y \) into Equation 2.

Substituting \( y \) in Equation 2:
\[
3x + 2(2x + 3) = 12
\]

### Step 2: Simplify the equation

Now distribute \( 2 \) to both terms in the parentheses:
\[
3x + 4x + 6 = 12
\]

Combine like terms:
\[
7x + 6 = 12
\]

### Step 3: Solve for \(x\)

Subtract \( 6 \) from both sides:
\[
7x = 12 - 6
\]
\[
7x = 6
\]

Now, divide by \( 7 \):
\[
x = \frac{6}{7}
\]

### Step 4: Substitute back to find \(y\)

Now that we have \( x \), we substitute it back into Equation 1 to find \( y \):
\[
y = 2\left(\frac{6}{7}\right) + 3
\]

Calculate \(y\):
\[
y = \frac{12}{7} + 3
\]
To combine these, convert \(3\) to a fraction with a denominator of \(7\):
\[
y = \frac{12}{7} + \frac{21}{7} = \frac{12 + 21}{7} = \frac{33}{7}
\]

### Final answer

The solution to the system of equations is:
\[
x = \frac{6}{7} \quad \text{and} \quad y = \frac{33}{7}
\]

Thus, the solution as an ordered pair is:
\[
\left( \frac{6}{7}, \frac{33}{7} \right)
\]