This system has one solution.
y=8x−14
y=x^2+4x−10
What is the y-coordinate of the solution?
y = 8x - 14
y = x^2 + 4x - 10
From these equations, you get:
8x - 14 = x^2 + 4x - 10
4x - 4 = x^2
x^2 - 4x + 4 = 0
(x - 2)^2 = 0
x - 2 = 0
x = 2
So, the x-coordinate of the solution is 2.
You can simply plus this into either equation to find the y-coordinate.
y = 8x - 14
= 8(2) - 14
= 16 - 14
= 2
To find the y-coordinate of the solution, we need to solve the system of equations by finding the value of y that satisfies both equations.
The given equations are:
1) y = 8x - 14
2) y = x^2 + 4x - 10
To solve this system, we can substitute the value of y from equation 1 into equation 2, then solve for x.
Substituting y = 8x - 14 into equation 2, we get:
8x - 14 = x^2 + 4x - 10
Rearranging the equation to standard form:
x^2 + 4x - 8x + 4 = 14 - 10
x^2 - 4x + 4 = 4
x^2 - 4x = 0
x(x - 4) = 0
This equation can be satisfied when either x = 0 or x - 4 = 0. So the possible x-values are x = 0 or x = 4.
To find the corresponding y-values, we substitute these x-values back into either equation 1 or 2.
For x = 0:
y = 8(0) - 14
y = -14
For x = 4:
y = 8(4) - 14
y = 32 - 14
y = 18
So, the possible y-coordinates of the solution are y = -14 and y = 18.