Answer :
To solve the given system of equations:
[tex]\[ \begin{cases} x^2 + y^2 = 25 \\ x^2 + y^2 - 10x - 8y = -32 \end{cases} \][/tex]
we can use the method of substitution or elimination. Here, let's proceed by subtracting the first equation from the second to eliminate [tex]\( x^2 + y^2 \)[/tex]:
1. Rewrite the equations:
[tex]\[ \begin{cases} x^2 + y^2 = 25 \quad \text{(1)} \\ x^2 + y^2 - 10x - 8y = -32 \quad \text{(2)} \end{cases} \][/tex]
2. Subtract (1) from (2):
[tex]\[ (x^2 + y^2 - 10x - 8y) - (x^2 + y^2) = -32 - 25 \][/tex]
[tex]\[ -10x - 8y = -57 \][/tex]
3. Simplify:
[tex]\[ 10x + 8y = 57 \quad \text{(3)} \][/tex]
4. To make things simpler, we can divide equation (3) by 2:
[tex]\[ 5x + 4y = \frac{57}{2} \quad \text{(4)} \][/tex]
5. Now, we have two equations:
[tex]\[ \begin{cases} x^2 + y^2 = 25 \quad \text{(1)} \\ 5x + 4y = \frac{57}{2} \quad \text{(4)} \end{cases} \][/tex]
Let's solve equation (4) for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ 4y = \frac{57}{2} - 5x \][/tex]
[tex]\[ y = \frac{57}{8} - \frac{5x}{4} \quad \text{(5)} \][/tex]
6. Substitute equation (5) into (1):
[tex]\[ x^2 + \left(\frac{57}{8} - \frac{5x}{4}\right)^2 = 25 \][/tex]
7. Expand the squared term in the equation:
[tex]\[ x^2 + \left(\frac{57}{8}\right)^2 - \frac{2 \cdot 57 \cdot 5x}{32} + \left(\frac{5x}{4}\right)^2 = 25 \][/tex]
[tex]\[ x^2 + \frac{3249}{64} - \frac{570x}{32} + \frac{25x^2}{16} = 25 \][/tex]
8. Multiply through by 64 to clear the fractions:
[tex]\[ 64x^2 + 3249 - 1140x + 100x^2 = 1600 \][/tex]
9. Combine like terms:
[tex]\[ 164x^2 - 1140x + 3249 = 1600 \][/tex]
10. Subtract 1600 from both sides to set the equation to zero:
[tex]\[ 164x^2 - 1140x + 1649 = 0 \][/tex]
This is a quadratic equation in [tex]\( x \)[/tex], which can be solved using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 164 \)[/tex], [tex]\( b = -1140 \)[/tex], and [tex]\( c = 1649 \)[/tex].
Solving this quadratic equation yields:
[tex]\[ x_1 = \frac{285 - 2\sqrt{851}}{82}, \quad x_2 = \frac{285 + 2\sqrt{851}}{82} \][/tex]
Now, substitute these values of [tex]\( x \)[/tex] back into equation (5) to find the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x_1 = \frac{285 - 2\sqrt{851}}{82} \)[/tex]:
[tex]\[ y_1 = \frac{57}{8} - \frac{5 \left( \frac{285 - 2\sqrt{851}}{82} \right)}{4} \][/tex]
Solving for [tex]\( y_1 \)[/tex], we get:
[tex]\[ y_1 = \frac{5\sqrt{851}}{82} + \frac{114}{41} \][/tex]
For [tex]\( x_2 = \frac{285 + 2\sqrt{851}}{82} \)[/tex]:
[tex]\[ y_2 = \frac{57}{8} - \frac{5 \left( \frac{285 + 2\sqrt{851}}{82} \right)}{4} \][/tex]
Solving for [tex]\( y_2 \)[/tex], we get:
[tex]\[ y_2 = \frac{114}{41} - \frac{5\sqrt{851}}{82} \][/tex]
Thus, the solutions to the system of equations are:
[tex]\[ \left( \frac{285 - 2\sqrt{851}}{82}, \frac{5\sqrt{851}}{82} + \frac{114}{41} \right) \quad \text{and} \quad \left( \frac{285 + 2\sqrt{851}}{82}, \frac{114}{41} - \frac{5\sqrt{851}}{82} \right) \][/tex]
[tex]\[ \begin{cases} x^2 + y^2 = 25 \\ x^2 + y^2 - 10x - 8y = -32 \end{cases} \][/tex]
we can use the method of substitution or elimination. Here, let's proceed by subtracting the first equation from the second to eliminate [tex]\( x^2 + y^2 \)[/tex]:
1. Rewrite the equations:
[tex]\[ \begin{cases} x^2 + y^2 = 25 \quad \text{(1)} \\ x^2 + y^2 - 10x - 8y = -32 \quad \text{(2)} \end{cases} \][/tex]
2. Subtract (1) from (2):
[tex]\[ (x^2 + y^2 - 10x - 8y) - (x^2 + y^2) = -32 - 25 \][/tex]
[tex]\[ -10x - 8y = -57 \][/tex]
3. Simplify:
[tex]\[ 10x + 8y = 57 \quad \text{(3)} \][/tex]
4. To make things simpler, we can divide equation (3) by 2:
[tex]\[ 5x + 4y = \frac{57}{2} \quad \text{(4)} \][/tex]
5. Now, we have two equations:
[tex]\[ \begin{cases} x^2 + y^2 = 25 \quad \text{(1)} \\ 5x + 4y = \frac{57}{2} \quad \text{(4)} \end{cases} \][/tex]
Let's solve equation (4) for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ 4y = \frac{57}{2} - 5x \][/tex]
[tex]\[ y = \frac{57}{8} - \frac{5x}{4} \quad \text{(5)} \][/tex]
6. Substitute equation (5) into (1):
[tex]\[ x^2 + \left(\frac{57}{8} - \frac{5x}{4}\right)^2 = 25 \][/tex]
7. Expand the squared term in the equation:
[tex]\[ x^2 + \left(\frac{57}{8}\right)^2 - \frac{2 \cdot 57 \cdot 5x}{32} + \left(\frac{5x}{4}\right)^2 = 25 \][/tex]
[tex]\[ x^2 + \frac{3249}{64} - \frac{570x}{32} + \frac{25x^2}{16} = 25 \][/tex]
8. Multiply through by 64 to clear the fractions:
[tex]\[ 64x^2 + 3249 - 1140x + 100x^2 = 1600 \][/tex]
9. Combine like terms:
[tex]\[ 164x^2 - 1140x + 3249 = 1600 \][/tex]
10. Subtract 1600 from both sides to set the equation to zero:
[tex]\[ 164x^2 - 1140x + 1649 = 0 \][/tex]
This is a quadratic equation in [tex]\( x \)[/tex], which can be solved using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 164 \)[/tex], [tex]\( b = -1140 \)[/tex], and [tex]\( c = 1649 \)[/tex].
Solving this quadratic equation yields:
[tex]\[ x_1 = \frac{285 - 2\sqrt{851}}{82}, \quad x_2 = \frac{285 + 2\sqrt{851}}{82} \][/tex]
Now, substitute these values of [tex]\( x \)[/tex] back into equation (5) to find the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x_1 = \frac{285 - 2\sqrt{851}}{82} \)[/tex]:
[tex]\[ y_1 = \frac{57}{8} - \frac{5 \left( \frac{285 - 2\sqrt{851}}{82} \right)}{4} \][/tex]
Solving for [tex]\( y_1 \)[/tex], we get:
[tex]\[ y_1 = \frac{5\sqrt{851}}{82} + \frac{114}{41} \][/tex]
For [tex]\( x_2 = \frac{285 + 2\sqrt{851}}{82} \)[/tex]:
[tex]\[ y_2 = \frac{57}{8} - \frac{5 \left( \frac{285 + 2\sqrt{851}}{82} \right)}{4} \][/tex]
Solving for [tex]\( y_2 \)[/tex], we get:
[tex]\[ y_2 = \frac{114}{41} - \frac{5\sqrt{851}}{82} \][/tex]
Thus, the solutions to the system of equations are:
[tex]\[ \left( \frac{285 - 2\sqrt{851}}{82}, \frac{5\sqrt{851}}{82} + \frac{114}{41} \right) \quad \text{and} \quad \left( \frac{285 + 2\sqrt{851}}{82}, \frac{114}{41} - \frac{5\sqrt{851}}{82} \right) \][/tex]
