Answer :
To solve the system of equations:
[tex]\[ \begin{aligned} 1. & \quad x^2 + y^2 = 20 \\ 2. & \quad 3x - y = 2 \end{aligned} \][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
### Step-by-Step Solution
1. Isolate [tex]\( y \)[/tex] in the second equation:
Starting with the second equation [tex]\(3x - y = 2\)[/tex], we can solve for [tex]\(y\)[/tex]:
[tex]\[ 3x - y = 2 \implies y = 3x - 2 \][/tex]
2. Substitute [tex]\( y \)[/tex] into the first equation:
Now that we have [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex], substitute [tex]\( y = 3x - 2 \)[/tex] into the first equation [tex]\(x^2 + y^2 = 20\)[/tex]:
[tex]\[ x^2 + (3x - 2)^2 = 20 \][/tex]
3. Expand and simplify:
Expand the square term and simplify the equation:
[tex]\[ x^2 + (3x - 2)^2 = x^2 + (9x^2 - 12x + 4) = 20 \][/tex]
[tex]\[ x^2 + 9x^2 - 12x + 4 = 20 \][/tex]
[tex]\[ 10x^2 - 12x + 4 = 20 \][/tex]
4. Move all terms to one side of the equation to form a quadratic equation:
[tex]\[ 10x^2 - 12x + 4 - 20 = 0 \][/tex]
[tex]\[ 10x^2 - 12x - 16 = 0 \][/tex]
5. Simplify the quadratic equation:
Divide each term by 2 to simplify the equation:
[tex]\[ 5x^2 - 6x - 8 = 0 \][/tex]
6. Solve the quadratic equation using the quadratic formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\(5x^2 - 6x - 8 = 0\)[/tex], we have [tex]\(a = 5\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = -8\)[/tex]. Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(5)(-8)}}{2(5)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 + 160}}{10} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{196}}{10} \][/tex]
[tex]\[ x = \frac{6 \pm 14}{10} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{6 + 14}{10} = 2 \quad \text{and} \quad x = \frac{6 - 14}{10} = -0.8 \][/tex]
7. Find corresponding [tex]\( y \)[/tex] values:
Substitute [tex]\( x \)[/tex] back into the equation [tex]\( y = 3x - 2 \)[/tex]:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 3(2) - 2 = 6 - 2 = 4 \][/tex]
- For [tex]\( x = -0.8 \)[/tex]:
[tex]\[ y = 3(-0.8) - 2 = -2.4 - 2 = -4.4 \][/tex]
8. Verify the solutions in the original equations:
- For [tex]\( x = 2 \)[/tex] and [tex]\( y = 4 \)[/tex]:
[tex]\[ 2^2 + 4^2 = 4 + 16 = 20 \quad \text{(true)} \][/tex]
[tex]\[ 3(2) - 4 = 6 - 4 = 2 \quad \text{(true)} \][/tex]
- For [tex]\( x = -0.8 \)[/tex] and [tex]\( y = -4.4 \)[/tex]:
[tex]\[ (-0.8)^2 + (-4.4)^2 = 0.64 + 19.36 = 20 \quad \text{(true)} \][/tex]
[tex]\[ 3(-0.8) - (-4.4) = -2.4 + 4.4 = 2 \quad \text{(true)} \][/tex]
Both pairs [tex]\((2, 4)\)[/tex] and [tex]\((-0.8, -4.4)\)[/tex] satisfy the original equations.
### Final Answer
The solutions to the system of equations are:
[tex]\[ (x, y) = (2, 4) \quad \text{and} \quad (x, y) = (-0.8, -4.4) \][/tex]
[tex]\[ \begin{aligned} 1. & \quad x^2 + y^2 = 20 \\ 2. & \quad 3x - y = 2 \end{aligned} \][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
### Step-by-Step Solution
1. Isolate [tex]\( y \)[/tex] in the second equation:
Starting with the second equation [tex]\(3x - y = 2\)[/tex], we can solve for [tex]\(y\)[/tex]:
[tex]\[ 3x - y = 2 \implies y = 3x - 2 \][/tex]
2. Substitute [tex]\( y \)[/tex] into the first equation:
Now that we have [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex], substitute [tex]\( y = 3x - 2 \)[/tex] into the first equation [tex]\(x^2 + y^2 = 20\)[/tex]:
[tex]\[ x^2 + (3x - 2)^2 = 20 \][/tex]
3. Expand and simplify:
Expand the square term and simplify the equation:
[tex]\[ x^2 + (3x - 2)^2 = x^2 + (9x^2 - 12x + 4) = 20 \][/tex]
[tex]\[ x^2 + 9x^2 - 12x + 4 = 20 \][/tex]
[tex]\[ 10x^2 - 12x + 4 = 20 \][/tex]
4. Move all terms to one side of the equation to form a quadratic equation:
[tex]\[ 10x^2 - 12x + 4 - 20 = 0 \][/tex]
[tex]\[ 10x^2 - 12x - 16 = 0 \][/tex]
5. Simplify the quadratic equation:
Divide each term by 2 to simplify the equation:
[tex]\[ 5x^2 - 6x - 8 = 0 \][/tex]
6. Solve the quadratic equation using the quadratic formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\(5x^2 - 6x - 8 = 0\)[/tex], we have [tex]\(a = 5\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = -8\)[/tex]. Substitute these values into the quadratic formula:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(5)(-8)}}{2(5)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 + 160}}{10} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{196}}{10} \][/tex]
[tex]\[ x = \frac{6 \pm 14}{10} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{6 + 14}{10} = 2 \quad \text{and} \quad x = \frac{6 - 14}{10} = -0.8 \][/tex]
7. Find corresponding [tex]\( y \)[/tex] values:
Substitute [tex]\( x \)[/tex] back into the equation [tex]\( y = 3x - 2 \)[/tex]:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 3(2) - 2 = 6 - 2 = 4 \][/tex]
- For [tex]\( x = -0.8 \)[/tex]:
[tex]\[ y = 3(-0.8) - 2 = -2.4 - 2 = -4.4 \][/tex]
8. Verify the solutions in the original equations:
- For [tex]\( x = 2 \)[/tex] and [tex]\( y = 4 \)[/tex]:
[tex]\[ 2^2 + 4^2 = 4 + 16 = 20 \quad \text{(true)} \][/tex]
[tex]\[ 3(2) - 4 = 6 - 4 = 2 \quad \text{(true)} \][/tex]
- For [tex]\( x = -0.8 \)[/tex] and [tex]\( y = -4.4 \)[/tex]:
[tex]\[ (-0.8)^2 + (-4.4)^2 = 0.64 + 19.36 = 20 \quad \text{(true)} \][/tex]
[tex]\[ 3(-0.8) - (-4.4) = -2.4 + 4.4 = 2 \quad \text{(true)} \][/tex]
Both pairs [tex]\((2, 4)\)[/tex] and [tex]\((-0.8, -4.4)\)[/tex] satisfy the original equations.
### Final Answer
The solutions to the system of equations are:
[tex]\[ (x, y) = (2, 4) \quad \text{and} \quad (x, y) = (-0.8, -4.4) \][/tex]
