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Sagot :
Certainly! Let's solve the given system of equations using the elimination method. The system of equations is:
[tex]\[ \begin{array}{l} -3x + 2y = 9 \\ x + y = 12 \end{array} \][/tex]
### Step-by-Step Solution
1. Label the equations for reference:
[tex]\[ \begin{array}{l} -3x + 2y = 9 \quad \text{(Equation 1)} \\ x + y = 12 \quad \text{(Equation 2)} \end{array} \][/tex]
2. Eliminate one variable:
To eliminate [tex]\(x\)[/tex], we can multiply Equation 2 by 3 so that the coefficients of [tex]\(x\)[/tex] will be opposites. This will allow us to add the equations and eliminate [tex]\(x\)[/tex].
[tex]\[ 3(x + y) = 3 \cdot 12 \][/tex]
Simplifying, we get:
[tex]\[ 3x + 3y = 36 \quad \text{(Equation 3)} \][/tex]
3. Add Equation 1 and Equation 3:
[tex]\[ \begin{array}{rcl} -3x + 2y & = & 9 \\ 3x + 3y & = & 36 \\ \hline 0x + 5y & = & 45 \end{array} \][/tex]
Simplifying, we get:
[tex]\[ 5y = 45 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{45}{5} = 9 \][/tex]
4. Substitute [tex]\(y\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex]:
We can use Equation 2:
[tex]\[ x + y = 12 \][/tex]
Substituting [tex]\(y = 9\)[/tex]:
[tex]\[ x + 9 = 12 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = 12 - 9 = 3 \][/tex]
### Solution
The solution to the system of equations is [tex]\((x, y) = (3, 9)\)[/tex].
So, the correct answer is:
[tex]\[ (3, 9) \][/tex]
Therefore, among the given choices:
- [tex]\((-3, 0)\)[/tex]
- [tex]\((1, 6)\)[/tex]
- [tex]\((3, 9)\)[/tex]
- [tex]\((5, 7)\)[/tex]
The correct solution is [tex]\((3, 9)\)[/tex].
[tex]\[ \begin{array}{l} -3x + 2y = 9 \\ x + y = 12 \end{array} \][/tex]
### Step-by-Step Solution
1. Label the equations for reference:
[tex]\[ \begin{array}{l} -3x + 2y = 9 \quad \text{(Equation 1)} \\ x + y = 12 \quad \text{(Equation 2)} \end{array} \][/tex]
2. Eliminate one variable:
To eliminate [tex]\(x\)[/tex], we can multiply Equation 2 by 3 so that the coefficients of [tex]\(x\)[/tex] will be opposites. This will allow us to add the equations and eliminate [tex]\(x\)[/tex].
[tex]\[ 3(x + y) = 3 \cdot 12 \][/tex]
Simplifying, we get:
[tex]\[ 3x + 3y = 36 \quad \text{(Equation 3)} \][/tex]
3. Add Equation 1 and Equation 3:
[tex]\[ \begin{array}{rcl} -3x + 2y & = & 9 \\ 3x + 3y & = & 36 \\ \hline 0x + 5y & = & 45 \end{array} \][/tex]
Simplifying, we get:
[tex]\[ 5y = 45 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{45}{5} = 9 \][/tex]
4. Substitute [tex]\(y\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex]:
We can use Equation 2:
[tex]\[ x + y = 12 \][/tex]
Substituting [tex]\(y = 9\)[/tex]:
[tex]\[ x + 9 = 12 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = 12 - 9 = 3 \][/tex]
### Solution
The solution to the system of equations is [tex]\((x, y) = (3, 9)\)[/tex].
So, the correct answer is:
[tex]\[ (3, 9) \][/tex]
Therefore, among the given choices:
- [tex]\((-3, 0)\)[/tex]
- [tex]\((1, 6)\)[/tex]
- [tex]\((3, 9)\)[/tex]
- [tex]\((5, 7)\)[/tex]
The correct solution is [tex]\((3, 9)\)[/tex].
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