Question

Solve the system of equations by substitution.
y=x^2-2x-7
y=-2x^2+6x-4

Answer 1

To solve the system of equations by substitution, we start by taking the expression for y from the first equation and substituting it into the second equation. This will give us an equation with only one variable, x, which we can solve. Here are the steps to follow:

The first equation is:
y = x^2 - 2x - 7

The second equation is:
y = -2x^2 + 6x - 4

Substituting the expression for y from the first equation into the second gives us:
x^2 - 2x - 7 = -2x^2 + 6x - 4

Now, we need to solve for x by combining like terms and moving everything to one side of the equation:
x^2 + 2x^2 - 2x - 6x - 7 + 4 = 0
3x^2 - 8x - 3 = 0

The equation 3x^2 - 8x - 3 = 0 is a quadratic equation, which we can solve using the quadratic formula, factoring, or completing the square. In this case, it looks like it doesn't factor easily, so let's use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

For the equation 3x^2 - 8x - 3 = 0, our coefficients are:
a = 3
b = -8
c = -3

Now we plug these values into the quadratic formula:
x = (-(-8) ± sqrt((-8)^2 - 4 * 3 * (-3))) / (2 * 3)
x = (8 ± sqrt(64 + 36)) / 6
x = (8 ± sqrt(100)) / 6
x = (8 ± 10) / 6

This gives us two possible solutions for x:
x = (8 + 10) / 6
x = 18 / 6
x = 3

and

x = (8 - 10) / 6
x = -2 / 6
x = -1/3

Now we can substitute each value of x back into either of the original equations to find the corresponding y value.

For x = 3:
y = 3^2 - 2*3 - 7
y = 9 - 6 - 7
y = 3 - 7
y = -4

The coordinate for x = 3 is (3, -4).

For x = -1/3:
y = (-1/3)^2 - 2*(-1/3) - 7
y = 1/9 + 2/3 - 7
y = 1/9 + 6/9 - 63/9
y = 7/9 - 63/9
y = -56/9

The coordinate for x = -1/3 is (-1/3, -56/9).

In conclusion, the system of equations has two solutions:
(3, -4) and (-1/3, -56/9).