Answer :
Let us solve the system of equations step-by-step.
The system of equations is:
[tex]\[ \begin{aligned} x & = 12 - y \quad \text{(Equation 1)} \\ 2x + 3y & = 29 \quad \text{(Equation 2)} \end{aligned} \][/tex]
1. Substitute Equation 1 into Equation 2:
From Equation 1, we know that:
[tex]\[ x = 12 - y \][/tex]
Substitute [tex]\( x = 12 - y \)[/tex] into Equation 2:
[tex]\[ 2(12 - y) + 3y = 29 \][/tex]
2. Simplify the resulting equation:
Distribute the 2:
[tex]\[ 24 - 2y + 3y = 29 \][/tex]
Combine like terms ([tex]\(-2y + 3y\)[/tex]):
[tex]\[ 24 + y = 29 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ y = 29 - 24 \][/tex]
[tex]\[ y = 5 \][/tex]
3. Substitute [tex]\( y = 5 \)[/tex] back into Equation 1 to find [tex]\( x \)[/tex]:
Using Equation 1:
[tex]\[ x = 12 - y \][/tex]
Substituting [tex]\( y = 5 \)[/tex]:
[tex]\[ x = 12 - 5 \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 7, \quad y = 5 \][/tex]
Thus, the correct answer is:
C. [tex]\( x = 7, y = 5 \)[/tex]
The system of equations is:
[tex]\[ \begin{aligned} x & = 12 - y \quad \text{(Equation 1)} \\ 2x + 3y & = 29 \quad \text{(Equation 2)} \end{aligned} \][/tex]
1. Substitute Equation 1 into Equation 2:
From Equation 1, we know that:
[tex]\[ x = 12 - y \][/tex]
Substitute [tex]\( x = 12 - y \)[/tex] into Equation 2:
[tex]\[ 2(12 - y) + 3y = 29 \][/tex]
2. Simplify the resulting equation:
Distribute the 2:
[tex]\[ 24 - 2y + 3y = 29 \][/tex]
Combine like terms ([tex]\(-2y + 3y\)[/tex]):
[tex]\[ 24 + y = 29 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ y = 29 - 24 \][/tex]
[tex]\[ y = 5 \][/tex]
3. Substitute [tex]\( y = 5 \)[/tex] back into Equation 1 to find [tex]\( x \)[/tex]:
Using Equation 1:
[tex]\[ x = 12 - y \][/tex]
Substituting [tex]\( y = 5 \)[/tex]:
[tex]\[ x = 12 - 5 \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 7, \quad y = 5 \][/tex]
Thus, the correct answer is:
C. [tex]\( x = 7, y = 5 \)[/tex]