Answer :
To solve the system of equations:
1. [tex]\(x^2 + y^2 = 25\)[/tex]
2. [tex]\(x^2 + y^2 - 10x - 8y = -32\)[/tex]
Let's start by simplifying and solving these equations step-by-step.
Step 1: Organize the equations
First, observe the two given equations:
[tex]\[ x^2 + y^2 = 25 \tag{1} \][/tex]
[tex]\[ x^2 + y^2 - 10x - 8y = -32 \tag{2} \][/tex]
Step 2: Subtract Equation (1) from Equation (2)
Subtract Equation (1) from Equation (2):
[tex]\[ (x^2 + y^2 - 10x - 8y) - (x^2 + y^2) = -32 - 25 \][/tex]
[tex]\[ -10x - 8y = -57 \][/tex]
[tex]\[ 10x + 8y = 57 \][/tex]
Step 3: Solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex] (or vice versa)
From [tex]\(10x + 8y = 57\)[/tex], solve for [tex]\(x\)[/tex]:
[tex]\[ 10x + 8y = 57 \][/tex]
[tex]\[ x = \frac{57 - 8y}{10} \][/tex]
Step 4: Substitute back into Equation (1)
Substitute [tex]\(x\)[/tex] from the above equation back into Equation (1):
[tex]\[ \left(\frac{57 - 8y}{10}\right)^2 + y^2 = 25 \][/tex]
Square [tex]\(\left(\frac{57 - 8y}{10}\right)\)[/tex] to get:
[tex]\[ \frac{(57 - 8y)^2}{100} + y^2 = 25 \][/tex]
[tex]\[ \frac{3249 - 2 \cdot 57 \cdot 8y + 64y^2}{100} + y^2 = 25 \][/tex]
[tex]\[ \frac{3249 - 912y + 64y^2}{100} + y^2 = 25 \][/tex]
Step 5: Multiply through by 100 to clear the fraction
[tex]\[ 3249 - 912y + 64y^2 + 100y^2 = 2500 \][/tex]
[tex]\[ 164y^2 - 912y + 3249 = 2500 \][/tex]
[tex]\[ 164y^2 - 912y + 749 = 0 \][/tex]
Step 6: Solve the quadratic equation for [tex]\(y\)[/tex]
This is a standard quadratic equation of the form [tex]\(ay^2 + by + c = 0\)[/tex], where:
[tex]\[ a = 164 \][/tex]
[tex]\[ b = -912 \][/tex]
[tex]\[ c = 749 \][/tex]
Using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{912 \pm \sqrt{912^2 - 4 \cdot 164 \cdot 749}}{2 \cdot 164} \][/tex]
[tex]\[ y = \frac{912 \pm \sqrt{831744 - 491344}}{328} \][/tex]
[tex]\[ y = \frac{912 \pm \sqrt{340400}}{328} \][/tex]
[tex]\[ y = \frac{912 \pm 583.459}{328} \][/tex]
This gives us two solutions for [tex]\(y\)[/tex]:
[tex]\[ y_1 = \frac{912 + 583.459}{328} = \frac{1495.459}{328} \approx 4.561 \][/tex]
[tex]\[ y_2 = \frac{912 - 583.459}{328} = \frac{328.541}{328} \approx 1.002 \][/tex]
Step 7: Substitute [tex]\(y_1\)[/tex] and [tex]\(y_2\)[/tex] back to find [tex]\(x\)[/tex]
For [tex]\(y_1 \approx 4.561\)[/tex]:
[tex]\[ x = \frac{57 - 8(4.561)}{10} \][/tex]
[tex]\[ x \approx \frac{57 - 36.488}{10} \][/tex]
[tex]\[ x \approx \frac{20.512}{10} \approx 2.051 \][/tex]
For [tex]\(y_2 \approx 1.002\)[/tex]:
[tex]\[ x = \frac{57 - 8(1.002)}{10} \][/tex]
[tex]\[ x \approx \frac{57 - 8.016}{10} \][/tex]
[tex]\[ x \approx \frac{48.984}{10} \approx 4.898 \][/tex]
Thus, the solutions to the system of equations are approximately [tex]\((x_1, y_1) \approx (2.051, 4.561)\)[/tex] and [tex]\((x_2, y_2) \approx (4.898, 1.002)\)[/tex].
The precise solutions, based on the given numerical result, are:
[tex]\[ \left( \frac{285}{82} - \frac{2\sqrt{851}}{41}, \frac{5\sqrt{851}}{82} + \frac{114}{41} \right), \quad \left( \frac{2\sqrt{851}}{41} + \frac{285}{82}, \frac{114}{41} - \frac{5\sqrt{851}}{82} \right) \][/tex]
1. [tex]\(x^2 + y^2 = 25\)[/tex]
2. [tex]\(x^2 + y^2 - 10x - 8y = -32\)[/tex]
Let's start by simplifying and solving these equations step-by-step.
Step 1: Organize the equations
First, observe the two given equations:
[tex]\[ x^2 + y^2 = 25 \tag{1} \][/tex]
[tex]\[ x^2 + y^2 - 10x - 8y = -32 \tag{2} \][/tex]
Step 2: Subtract Equation (1) from Equation (2)
Subtract Equation (1) from Equation (2):
[tex]\[ (x^2 + y^2 - 10x - 8y) - (x^2 + y^2) = -32 - 25 \][/tex]
[tex]\[ -10x - 8y = -57 \][/tex]
[tex]\[ 10x + 8y = 57 \][/tex]
Step 3: Solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex] (or vice versa)
From [tex]\(10x + 8y = 57\)[/tex], solve for [tex]\(x\)[/tex]:
[tex]\[ 10x + 8y = 57 \][/tex]
[tex]\[ x = \frac{57 - 8y}{10} \][/tex]
Step 4: Substitute back into Equation (1)
Substitute [tex]\(x\)[/tex] from the above equation back into Equation (1):
[tex]\[ \left(\frac{57 - 8y}{10}\right)^2 + y^2 = 25 \][/tex]
Square [tex]\(\left(\frac{57 - 8y}{10}\right)\)[/tex] to get:
[tex]\[ \frac{(57 - 8y)^2}{100} + y^2 = 25 \][/tex]
[tex]\[ \frac{3249 - 2 \cdot 57 \cdot 8y + 64y^2}{100} + y^2 = 25 \][/tex]
[tex]\[ \frac{3249 - 912y + 64y^2}{100} + y^2 = 25 \][/tex]
Step 5: Multiply through by 100 to clear the fraction
[tex]\[ 3249 - 912y + 64y^2 + 100y^2 = 2500 \][/tex]
[tex]\[ 164y^2 - 912y + 3249 = 2500 \][/tex]
[tex]\[ 164y^2 - 912y + 749 = 0 \][/tex]
Step 6: Solve the quadratic equation for [tex]\(y\)[/tex]
This is a standard quadratic equation of the form [tex]\(ay^2 + by + c = 0\)[/tex], where:
[tex]\[ a = 164 \][/tex]
[tex]\[ b = -912 \][/tex]
[tex]\[ c = 749 \][/tex]
Using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], solve for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{912 \pm \sqrt{912^2 - 4 \cdot 164 \cdot 749}}{2 \cdot 164} \][/tex]
[tex]\[ y = \frac{912 \pm \sqrt{831744 - 491344}}{328} \][/tex]
[tex]\[ y = \frac{912 \pm \sqrt{340400}}{328} \][/tex]
[tex]\[ y = \frac{912 \pm 583.459}{328} \][/tex]
This gives us two solutions for [tex]\(y\)[/tex]:
[tex]\[ y_1 = \frac{912 + 583.459}{328} = \frac{1495.459}{328} \approx 4.561 \][/tex]
[tex]\[ y_2 = \frac{912 - 583.459}{328} = \frac{328.541}{328} \approx 1.002 \][/tex]
Step 7: Substitute [tex]\(y_1\)[/tex] and [tex]\(y_2\)[/tex] back to find [tex]\(x\)[/tex]
For [tex]\(y_1 \approx 4.561\)[/tex]:
[tex]\[ x = \frac{57 - 8(4.561)}{10} \][/tex]
[tex]\[ x \approx \frac{57 - 36.488}{10} \][/tex]
[tex]\[ x \approx \frac{20.512}{10} \approx 2.051 \][/tex]
For [tex]\(y_2 \approx 1.002\)[/tex]:
[tex]\[ x = \frac{57 - 8(1.002)}{10} \][/tex]
[tex]\[ x \approx \frac{57 - 8.016}{10} \][/tex]
[tex]\[ x \approx \frac{48.984}{10} \approx 4.898 \][/tex]
Thus, the solutions to the system of equations are approximately [tex]\((x_1, y_1) \approx (2.051, 4.561)\)[/tex] and [tex]\((x_2, y_2) \approx (4.898, 1.002)\)[/tex].
The precise solutions, based on the given numerical result, are:
[tex]\[ \left( \frac{285}{82} - \frac{2\sqrt{851}}{41}, \frac{5\sqrt{851}}{82} + \frac{114}{41} \right), \quad \left( \frac{2\sqrt{851}}{41} + \frac{285}{82}, \frac{114}{41} - \frac{5\sqrt{851}}{82} \right) \][/tex]
