Question

10 solve the system of equations using elimination.
3x - 2y = 12
-x + y = -3

Explanation:

Step1: Label the equations

Let the first equation be \( 3x - 2y = 12 \) (Equation 1) and the second equation be \( -x + y = -3 \) (Equation 2).

Step2: Eliminate a variable

Multiply Equation 2 by 2 to make the coefficients of \( y \) opposite:
\( 2\times(-x + y)=2\times(-3) \)
\( -2x + 2y = -6 \) (Equation 3)

Step3: Add Equation 1 and Equation 3

Add \( 3x - 2y = 12 \) and \( -2x + 2y = -6 \):
\( (3x - 2y)+(-2x + 2y)=12+(-6) \)
\( 3x - 2y - 2x + 2y = 6 \)
Simplify: \( x = 6 \)

Step4: Substitute \( x \) into Equation 2

Substitute \( x = 6 \) into \( -x + y = -3 \):
\( -6 + y = -3 \)

Step5: Solve for \( y \)

Add 6 to both sides:
\( y = -3 + 6 \)
\( y = 3 \)

Answer:

The solution to the system is \( x = 6 \), \( y = 3 \)