Solve the following system of equations algebraically:y, equals, x, squared, plus, 2, x, minus, 29y=x 2 +2x−29y, equals, minus, 3, x, plus, 7y=−3x+7

The system of equations is:

1. y = x^2 + 2x - 29
2. y = -3x + 7

To solve this system algebraically, we can use the method of substitution or elimination. Here, since y is already isolated in both equations, we can set the two expressions for y equal to each other and solve for x:

x^2 + 2x - 29 = -3x + 7

Rearrange the equation to set it equal to zero:

x^2 + 2x + 3x - 29 - 7 = 0

Combine like terms:

x^2 + 5x - 36 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a):

x = [-5 ± sqrt((5)^2 - 41(-36))] / (2*1) x = [-5 ± sqrt(25 + 144)] / 2 x = [-5 ± sqrt(169)] / 2 x = [-5 ± 13] / 2

So the solutions for x are x = 4 and x = -9.

Substitute x = 4 into the second equation to find the corresponding y:

y = -3*4 + 7 = -12 + 7 = -5

Substitute x = -9 into the second equation to find the corresponding y:

y = -3*(-9) + 7 = 27 + 7 = 34

So the solutions to the system of equations are (4, -5) and (-9, 34).