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Sagot :
To solve the system of simultaneous equations:
[tex]\[ \begin{aligned} x & = 7 - 3y \\ x^2 - y^2 & = 39 \end{aligned} \][/tex]
we'll proceed step-by-step:
### Step 1: Substitute [tex]\( x \)[/tex] from the first equation into the second equation
Given:
[tex]\[ x = 7 - 3y \][/tex]
Substitute [tex]\( x = 7 - 3y \)[/tex] into the second equation [tex]\( x^2 - y^2 = 39 \)[/tex]:
[tex]\[ (7 - 3y)^2 - y^2 = 39 \][/tex]
### Step 2: Expand and simplify
Expand [tex]\( (7 - 3y)^2 \)[/tex]:
[tex]\[ (7 - 3y)^2 = 49 - 42y + 9y^2 \][/tex]
Now substitute this back in:
[tex]\[ 49 - 42y + 9y^2 - y^2 = 39 \][/tex]
Simplify the equation:
[tex]\[ 49 - 42y + 8y^2 = 39 \][/tex]
### Step 3: Combine like terms and solve for [tex]\( y \)[/tex]:
[tex]\[ 8y^2 - 42y + 49 - 39 = 0 \][/tex]
[tex]\[ 8y^2 - 42y + 10 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve [tex]\( 8y^2 - 42y + 10 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 8 \)[/tex], [tex]\( b = -42 \)[/tex], and [tex]\( c = 10 \)[/tex].
So,
[tex]\[ y = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 8 \cdot 10}}{2 \cdot 8} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1764 - 320}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1444}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm 38}{16} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{42 + 38}{16} = \frac{80}{16} = 5 \][/tex]
or
[tex]\[ y = \frac{42 - 38}{16} = \frac{4}{16} = \frac{1}{4} \][/tex]
### Step 5: Substitute [tex]\( y \)[/tex] back to find [tex]\( x \)[/tex]:
1. When [tex]\( y = 5 \)[/tex]:
[tex]\[ x = 7 - 3 \cdot 5 = 7 - 15 = -8 \][/tex]
2. When [tex]\( y = \frac{1}{4} \)[/tex]:
[tex]\[ x = 7 - 3 \cdot \frac{1}{4} = 7 - \frac{3}{4} = \frac{28}{4} - \frac{3}{4} = \frac{25}{4} \][/tex]
### Step 6: List the solutions
The solutions for the system of equations are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
Thus, the correct solutions are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
[tex]\[ \begin{aligned} x & = 7 - 3y \\ x^2 - y^2 & = 39 \end{aligned} \][/tex]
we'll proceed step-by-step:
### Step 1: Substitute [tex]\( x \)[/tex] from the first equation into the second equation
Given:
[tex]\[ x = 7 - 3y \][/tex]
Substitute [tex]\( x = 7 - 3y \)[/tex] into the second equation [tex]\( x^2 - y^2 = 39 \)[/tex]:
[tex]\[ (7 - 3y)^2 - y^2 = 39 \][/tex]
### Step 2: Expand and simplify
Expand [tex]\( (7 - 3y)^2 \)[/tex]:
[tex]\[ (7 - 3y)^2 = 49 - 42y + 9y^2 \][/tex]
Now substitute this back in:
[tex]\[ 49 - 42y + 9y^2 - y^2 = 39 \][/tex]
Simplify the equation:
[tex]\[ 49 - 42y + 8y^2 = 39 \][/tex]
### Step 3: Combine like terms and solve for [tex]\( y \)[/tex]:
[tex]\[ 8y^2 - 42y + 49 - 39 = 0 \][/tex]
[tex]\[ 8y^2 - 42y + 10 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve [tex]\( 8y^2 - 42y + 10 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 8 \)[/tex], [tex]\( b = -42 \)[/tex], and [tex]\( c = 10 \)[/tex].
So,
[tex]\[ y = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 8 \cdot 10}}{2 \cdot 8} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1764 - 320}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1444}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm 38}{16} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{42 + 38}{16} = \frac{80}{16} = 5 \][/tex]
or
[tex]\[ y = \frac{42 - 38}{16} = \frac{4}{16} = \frac{1}{4} \][/tex]
### Step 5: Substitute [tex]\( y \)[/tex] back to find [tex]\( x \)[/tex]:
1. When [tex]\( y = 5 \)[/tex]:
[tex]\[ x = 7 - 3 \cdot 5 = 7 - 15 = -8 \][/tex]
2. When [tex]\( y = \frac{1}{4} \)[/tex]:
[tex]\[ x = 7 - 3 \cdot \frac{1}{4} = 7 - \frac{3}{4} = \frac{28}{4} - \frac{3}{4} = \frac{25}{4} \][/tex]
### Step 6: List the solutions
The solutions for the system of equations are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
Thus, the correct solutions are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
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