Use the elimination method to solve the linear-quadratic system y=x^2+10 , y=−7x−2 .

Answer 1

To solve the linear-quadratic system using the elimination method, we first need to eliminate one variable.

We'll focus on eliminating y. The equation y = x^2 + 10 is already in the desired form, so we'll leave it as it is.

Next, we'll rewrite the equation y = -7x - 2 in the form x = (y + 2) / -7 to eliminate y.

Now, we have the following two equations:
1) y = x^2 + 10
2) x = (y + 2) / -7

To eliminate y, we can substitute equation 2 into equation 1 since they both represent y.

Substituting equation 2 into equation 1:
x^2 + 10 = ((x + 2) / -7) + 10

To get rid of the fraction, we can multiply the entire equation by -7:
-7(x^2 + 10) = x + 2 + (-70)

Simplifying the equation:
-7x^2 - 70 = x - 68

Rearranging the equation:
7x^2 + x - 2 = 0

Now, we have a quadratic equation. To solve for x, we can factor or use the quadratic formula. In this case, factoring seems easier:
(7x - 1)(x + 2) = 0

Setting each factor equal to zero:
7x - 1 = 0 or x + 2 = 0

Solving for x:
7x = 1 or x = -2

From the first equation:
x = 1/7

From the second equation:
x = -2

Now, we substitute these values of x into either one of the original equations to solve for y.

Using y = x^2 + 10:

When x = 1/7:
y = (1/7)^2 + 10
y = 1/49 + 10
y = 10 + 1/49
y = 490/49 + 1/49
y = 491/49

When x = -2:
y = (-2)^2 + 10
y = 4 + 10
y = 14

Therefore, the solution to the linear-quadratic system is:
(x, y) = (1/7, 491/49) and (-2, 14)