To solve the system of equations using substitution, we can start with the two equations given:
- \( y = 2x + 3 \) (Equation 1)
- \( 3x + 2y = 12 \) (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: \[ \left( \frac{6}{7}, \frac{33}{7} \right) \]
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