Answer :
To solve the system of equations:
[tex]\[ \begin{array}{c} \sqrt{x^2 + y^2} = 25 \\ y - 2x = 5 \end{array} \][/tex]
we will follow these steps:
1. Simplify the First Equation:
The first equation can be simplified by squaring both sides:
[tex]\[ \sqrt{x^2 + y^2} = 25 \implies x^2 + y^2 = 625 \][/tex]
2. Express [tex]\( y \)[/tex] in Terms of [tex]\( x \)[/tex]:
From the second equation, we can express [tex]\( y \)[/tex] as:
[tex]\[ y - 2x = 5 \implies y = 2x + 5 \][/tex]
3. Substitute [tex]\( y \)[/tex] into the First Equation:
Substitute [tex]\( y = 2x + 5 \)[/tex] into [tex]\( x^2 + y^2 = 625 \)[/tex]:
[tex]\[ x^2 + (2x + 5)^2 = 625 \][/tex]
4. Expand and Simplify the Equation:
Expand [tex]\( (2x + 5)^2 \)[/tex]:
[tex]\[ x^2 + (2x + 5)^2 = x^2 + (4x^2 + 20x + 25) = 625 \implies x^2 + 4x^2 + 20x + 25 = 625 \][/tex]
Combine like terms:
[tex]\[ 5x^2 + 20x + 25 = 625 \][/tex]
Subtract 625 from both sides:
[tex]\[ 5x^2 + 20x + 25 - 625 = 0 \implies 5x^2 + 20x - 600 = 0 \][/tex]
Simplify by dividing the entire equation by 5:
[tex]\[ x^2 + 4x - 120 = 0 \][/tex]
5. Solve the Quadratic Equation:
Solve [tex]\( x^2 + 4x - 120 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -120 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 480}}{2} = \frac{-4 \pm \sqrt{496}}{2} = \frac{-4 \pm 4\sqrt{31}}{2} = -2 \pm 2\sqrt{31} \][/tex]
So, [tex]\( x = -2 + 2\sqrt{31} \)[/tex] or [tex]\( x = -2 - 2\sqrt{31} \)[/tex].
6. Find the Corresponding [tex]\( y \)[/tex] Values:
Using [tex]\( y = 2x + 5 \)[/tex]:
- For [tex]\( x = -2 + 2\sqrt{31} \)[/tex]:
[tex]\[ y = 2(-2 + 2\sqrt{31}) + 5 = -4 + 4\sqrt{31} + 5 = 1 + 4\sqrt{31} \][/tex]
- For [tex]\( x = -2 - 2\sqrt{31} \)[/tex]:
[tex]\[ y = 2(-2 - 2\sqrt{31}) + 5 = -4 - 4\sqrt{31} + 5 = 1 - 4\sqrt{31} \][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[ \left( -2 + 2\sqrt{31}, 1 + 4\sqrt{31} \right) \quad \text{and} \quad \left( -2 - 2\sqrt{31}, 1 - 4\sqrt{31} \right) \][/tex]
[tex]\[ \begin{array}{c} \sqrt{x^2 + y^2} = 25 \\ y - 2x = 5 \end{array} \][/tex]
we will follow these steps:
1. Simplify the First Equation:
The first equation can be simplified by squaring both sides:
[tex]\[ \sqrt{x^2 + y^2} = 25 \implies x^2 + y^2 = 625 \][/tex]
2. Express [tex]\( y \)[/tex] in Terms of [tex]\( x \)[/tex]:
From the second equation, we can express [tex]\( y \)[/tex] as:
[tex]\[ y - 2x = 5 \implies y = 2x + 5 \][/tex]
3. Substitute [tex]\( y \)[/tex] into the First Equation:
Substitute [tex]\( y = 2x + 5 \)[/tex] into [tex]\( x^2 + y^2 = 625 \)[/tex]:
[tex]\[ x^2 + (2x + 5)^2 = 625 \][/tex]
4. Expand and Simplify the Equation:
Expand [tex]\( (2x + 5)^2 \)[/tex]:
[tex]\[ x^2 + (2x + 5)^2 = x^2 + (4x^2 + 20x + 25) = 625 \implies x^2 + 4x^2 + 20x + 25 = 625 \][/tex]
Combine like terms:
[tex]\[ 5x^2 + 20x + 25 = 625 \][/tex]
Subtract 625 from both sides:
[tex]\[ 5x^2 + 20x + 25 - 625 = 0 \implies 5x^2 + 20x - 600 = 0 \][/tex]
Simplify by dividing the entire equation by 5:
[tex]\[ x^2 + 4x - 120 = 0 \][/tex]
5. Solve the Quadratic Equation:
Solve [tex]\( x^2 + 4x - 120 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -120 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 480}}{2} = \frac{-4 \pm \sqrt{496}}{2} = \frac{-4 \pm 4\sqrt{31}}{2} = -2 \pm 2\sqrt{31} \][/tex]
So, [tex]\( x = -2 + 2\sqrt{31} \)[/tex] or [tex]\( x = -2 - 2\sqrt{31} \)[/tex].
6. Find the Corresponding [tex]\( y \)[/tex] Values:
Using [tex]\( y = 2x + 5 \)[/tex]:
- For [tex]\( x = -2 + 2\sqrt{31} \)[/tex]:
[tex]\[ y = 2(-2 + 2\sqrt{31}) + 5 = -4 + 4\sqrt{31} + 5 = 1 + 4\sqrt{31} \][/tex]
- For [tex]\( x = -2 - 2\sqrt{31} \)[/tex]:
[tex]\[ y = 2(-2 - 2\sqrt{31}) + 5 = -4 - 4\sqrt{31} + 5 = 1 - 4\sqrt{31} \][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[ \left( -2 + 2\sqrt{31}, 1 + 4\sqrt{31} \right) \quad \text{and} \quad \left( -2 - 2\sqrt{31}, 1 - 4\sqrt{31} \right) \][/tex]
