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Sagot :
Certainly! Let's solve the given system of equations step by step. The system is:
[tex]\[ \begin{cases} 3a + b = 2 \quad \text{(Equation 1)} \\ 9a^2 - b = 7 \quad \text{(Equation 2)} \end{cases} \][/tex]
### Step 1: Express [tex]\( b \)[/tex] from Equation 1
From Equation 1, we can isolate [tex]\( b \)[/tex]:
[tex]\[ b = 2 - 3a \][/tex]
### Step 2: Substitute [tex]\( b \)[/tex] into Equation 2
Now, we substitute [tex]\( b = 2 - 3a \)[/tex] into Equation 2:
[tex]\[ 9a^2 - (2 - 3a) = 7 \][/tex]
### Step 3: Simplify the resulting equation
Simplify the left-hand side:
[tex]\[ 9a^2 - 2 + 3a = 7 \][/tex]
Move all terms to one side to form a standard quadratic equation:
[tex]\[ 9a^2 + 3a - 9 = 0 \][/tex]
Divide the entire equation by 3 to simplify:
[tex]\[ 3a^2 + a - 3 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve the quadratic equation [tex]\( 3a^2 + a - 3 = 0 \)[/tex], we use the quadratic formula:
[tex]\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 3 \cdot (-3) = 1 + 36 = 37 \][/tex]
[tex]\[ a = \frac{-1 \pm \sqrt{37}}{6} \][/tex]
So we get the solutions:
[tex]\[ a_1 = \frac{-1 + \sqrt{37}}{6} \][/tex]
[tex]\[ a_2 = \frac{-1 - \sqrt{37}}{6} \][/tex]
### Step 5: Find corresponding [tex]\( b \)[/tex] values
Using [tex]\( b = 2 - 3a \)[/tex], we substitute [tex]\( a_1 \)[/tex] and [tex]\( a_2 \)[/tex]:
For [tex]\( a_1 = \frac{-1 + \sqrt{37}}{6} \)[/tex]:
[tex]\[ b_1 = 2 - 3 \left( \frac{-1 + \sqrt{37}}{6} \right) = \frac{12 - 3(-1 + \sqrt{37})}{6} = \frac{12 + 3 - 3\sqrt{37}}{6} = \frac{15 - 3\sqrt{37}}{6} = \frac{5 - \sqrt{37}}{2} \][/tex]
For [tex]\( a_2 = \frac{-1 - \sqrt{37}}{6} \)[/tex]:
[tex]\[ b_2 = 2 - 3 \left( \frac{-1 - \sqrt{37}}{6} \right) = \frac{12 - 3(-1 - \sqrt{37})}{6} = \frac{12 + 3 + 3\sqrt{37}}{6} = \frac{15 + 3\sqrt{37}}{6} = \frac{5 + \sqrt{37}}{2} \][/tex]
### Conclusion
The solutions to the system of equations are:
[tex]\[ \left(a, b\right) = \left( \frac{-1 + \sqrt{37}}{6}, \frac{5 - \sqrt{37}}{2} \right) \][/tex]
and
[tex]\[ \left(a, b\right) = \left( \frac{-1 - \sqrt{37}}{6}, \frac{5 + \sqrt{37}}{2} \right) \][/tex]
Thus, the solutions are:
[tex]\[ \left( \frac{-1 + \sqrt{37}}{6}, \frac{5 - \sqrt{37}}{2} \right) \quad \text{and} \quad \left( \frac{-1 - \sqrt{37}}{6}, \frac{5 + \sqrt{37}}{2} \right) \][/tex]
[tex]\[ \begin{cases} 3a + b = 2 \quad \text{(Equation 1)} \\ 9a^2 - b = 7 \quad \text{(Equation 2)} \end{cases} \][/tex]
### Step 1: Express [tex]\( b \)[/tex] from Equation 1
From Equation 1, we can isolate [tex]\( b \)[/tex]:
[tex]\[ b = 2 - 3a \][/tex]
### Step 2: Substitute [tex]\( b \)[/tex] into Equation 2
Now, we substitute [tex]\( b = 2 - 3a \)[/tex] into Equation 2:
[tex]\[ 9a^2 - (2 - 3a) = 7 \][/tex]
### Step 3: Simplify the resulting equation
Simplify the left-hand side:
[tex]\[ 9a^2 - 2 + 3a = 7 \][/tex]
Move all terms to one side to form a standard quadratic equation:
[tex]\[ 9a^2 + 3a - 9 = 0 \][/tex]
Divide the entire equation by 3 to simplify:
[tex]\[ 3a^2 + a - 3 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve the quadratic equation [tex]\( 3a^2 + a - 3 = 0 \)[/tex], we use the quadratic formula:
[tex]\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 3 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 3 \cdot (-3) = 1 + 36 = 37 \][/tex]
[tex]\[ a = \frac{-1 \pm \sqrt{37}}{6} \][/tex]
So we get the solutions:
[tex]\[ a_1 = \frac{-1 + \sqrt{37}}{6} \][/tex]
[tex]\[ a_2 = \frac{-1 - \sqrt{37}}{6} \][/tex]
### Step 5: Find corresponding [tex]\( b \)[/tex] values
Using [tex]\( b = 2 - 3a \)[/tex], we substitute [tex]\( a_1 \)[/tex] and [tex]\( a_2 \)[/tex]:
For [tex]\( a_1 = \frac{-1 + \sqrt{37}}{6} \)[/tex]:
[tex]\[ b_1 = 2 - 3 \left( \frac{-1 + \sqrt{37}}{6} \right) = \frac{12 - 3(-1 + \sqrt{37})}{6} = \frac{12 + 3 - 3\sqrt{37}}{6} = \frac{15 - 3\sqrt{37}}{6} = \frac{5 - \sqrt{37}}{2} \][/tex]
For [tex]\( a_2 = \frac{-1 - \sqrt{37}}{6} \)[/tex]:
[tex]\[ b_2 = 2 - 3 \left( \frac{-1 - \sqrt{37}}{6} \right) = \frac{12 - 3(-1 - \sqrt{37})}{6} = \frac{12 + 3 + 3\sqrt{37}}{6} = \frac{15 + 3\sqrt{37}}{6} = \frac{5 + \sqrt{37}}{2} \][/tex]
### Conclusion
The solutions to the system of equations are:
[tex]\[ \left(a, b\right) = \left( \frac{-1 + \sqrt{37}}{6}, \frac{5 - \sqrt{37}}{2} \right) \][/tex]
and
[tex]\[ \left(a, b\right) = \left( \frac{-1 - \sqrt{37}}{6}, \frac{5 + \sqrt{37}}{2} \right) \][/tex]
Thus, the solutions are:
[tex]\[ \left( \frac{-1 + \sqrt{37}}{6}, \frac{5 - \sqrt{37}}{2} \right) \quad \text{and} \quad \left( \frac{-1 - \sqrt{37}}{6}, \frac{5 + \sqrt{37}}{2} \right) \][/tex]
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