Answer :
To solve the given system of equations using the substitution method, follow these steps:
Given the system of equations:
[tex]\[ \begin{array}{l} 3x + 2y = 13 \quad \text{(Equation 1)} \\ y = x - 1 \quad \text{(Equation 2)} \end{array} \][/tex]
1. Substitute the expression for \( y \) from Equation 2 into Equation 1.
From Equation 2, we know:
[tex]\[ y = x - 1 \][/tex]
Substitute \( y = x - 1 \) into Equation 1:
[tex]\[ 3x + 2(x - 1) = 13 \][/tex]
2. Simplify and solve for \( x \):
[tex]\[ 3x + 2x - 2 = 13 \][/tex]
Combine like terms:
[tex]\[ 5x - 2 = 13 \][/tex]
Add 2 to both sides of the equation:
[tex]\[ 5x = 15 \][/tex]
Divide both sides by 5:
[tex]\[ x = 3 \][/tex]
3. Substitute \( x = 3 \) back into Equation 2 to find \( y \):
[tex]\[ y = x - 1 \][/tex]
[tex]\[ y = 3 - 1 \][/tex]
[tex]\[ y = 2 \][/tex]
4. Verify the solution by substituting \( x = 3 \) and \( y = 2 \) into both original equations:
Check Equation 1:
[tex]\[ 3x + 2y = 13 \][/tex]
[tex]\[ 3(3) + 2(2) = 13 \][/tex]
[tex]\[ 9 + 4 = 13 \][/tex]
Check Equation 2:
[tex]\[ y = x - 1 \][/tex]
[tex]\[ 2 = 3 - 1 \][/tex]
Both equations are satisfied with \( x = 3 \) and \( y = 2 \).
Therefore, the solution to the system of equations is:
[tex]\[ \boxed{(3, 2)} \][/tex]
The correct answer is:
[tex]\[ \text{B. } (3, 2) \][/tex]
Given the system of equations:
[tex]\[ \begin{array}{l} 3x + 2y = 13 \quad \text{(Equation 1)} \\ y = x - 1 \quad \text{(Equation 2)} \end{array} \][/tex]
1. Substitute the expression for \( y \) from Equation 2 into Equation 1.
From Equation 2, we know:
[tex]\[ y = x - 1 \][/tex]
Substitute \( y = x - 1 \) into Equation 1:
[tex]\[ 3x + 2(x - 1) = 13 \][/tex]
2. Simplify and solve for \( x \):
[tex]\[ 3x + 2x - 2 = 13 \][/tex]
Combine like terms:
[tex]\[ 5x - 2 = 13 \][/tex]
Add 2 to both sides of the equation:
[tex]\[ 5x = 15 \][/tex]
Divide both sides by 5:
[tex]\[ x = 3 \][/tex]
3. Substitute \( x = 3 \) back into Equation 2 to find \( y \):
[tex]\[ y = x - 1 \][/tex]
[tex]\[ y = 3 - 1 \][/tex]
[tex]\[ y = 2 \][/tex]
4. Verify the solution by substituting \( x = 3 \) and \( y = 2 \) into both original equations:
Check Equation 1:
[tex]\[ 3x + 2y = 13 \][/tex]
[tex]\[ 3(3) + 2(2) = 13 \][/tex]
[tex]\[ 9 + 4 = 13 \][/tex]
Check Equation 2:
[tex]\[ y = x - 1 \][/tex]
[tex]\[ 2 = 3 - 1 \][/tex]
Both equations are satisfied with \( x = 3 \) and \( y = 2 \).
Therefore, the solution to the system of equations is:
[tex]\[ \boxed{(3, 2)} \][/tex]
The correct answer is:
[tex]\[ \text{B. } (3, 2) \][/tex]
