Solve the following system of equations algebraically: y= x^2 −19x+58 y, y= −3x−5

Answer 1

To solve this system of equations algebraically, we can set the two expressions for y equal to each other and solve for x:

x^2 −19x + 58 = −3x − 5

Rearranging the terms:

x^2 − 16x + 63 = 0

Now we need to factor this quadratic equation:

(x − 7)(x − 9) = 0

Setting each factor equal to zero:

x - 7 = 0 or x - 9 = 0

x = 7 or x = 9

Now that we have found the values of x, we can substitute them back into one of the original equations to find the corresponding values of y. Let's use y = x^2 − 19x + 58:

For x = 7:

y = 7^2 − 19(7) + 58
y = 49 - 133 + 58
y = -26 + 58
y = 32

So when x = 7, y = 32.

For x = 9:

y = 9^2 − 19(9) + 58
y = 81 - 171 + 58
y = -90 + 58
y = -32

So when x = 9, y = -32.

Therefore, the solution to the system of equations is x = 7, y = 32 and x = 9, y = -32.