To solve the system of equations given by y = x² and y = 2x + 7, we can use substitution.
1. Since both equations equal y, we can set them equal to each other:
x² = 2x + 7
2. Next, rearrange the equation to form a standard quadratic equation:
x² - 2x - 7 = 0
3. Now, we can factor or use the quadratic formula to find the values of x. In this case, we’ll use the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
Where:
- a = 1 (the coefficient of x²)
- b = -2 (the coefficient of x)
- c = -7 (the constant term)
4. Substitute the values into the quadratic formula:
x = (2 ± √((-2)² - 4(1)(-7))) / (2(1))
5. Simplify inside the square root:
x = (2 ± √(4 + 28)) / 2
x = (2 ± √32) / 2
6. Notice that √32 can be simplified to 4√2, so:
x = (2 ± 4√2) / 2
7. This results in two solutions for x:
x = 1 + 2√2
x = 1 - 2√2
8. Now that we have the x values, we can find the corresponding y values by substituting each x back into either original equation. It’s usually easiest to use y = x²:
For x = 1 + 2√2:
y = (1 + 2√2)² = 1 + 4√2 + 8 = 9 + 4√2
For x = 1 – 2√2:
y = (1 - 2√2)² = 1 - 4√2 + 8 = 9 - 4√2
9. Therefore, the system of equations yields the solutions:
- (1 + 2√2, 9 + 4√2)
- (1 – 2√2, 9 – 4√2)
In conclusion, we solved the system of equations by substitution and derived two points where the parabolas intersect!