Final answer:
To solve the system of equations algebraically, we can rearrange the equations and use the quadratic formula to find the values of x that satisfy the equation. The solutions for x are 1/3 and -3/2. Substituting these values into one of the original equations will give us the corresponding values of y.
Step-by-step explanation:
To solve the system of equations y = 6x² + 5x - 13 and y=-2x - 10 algebraically, we can set the two equations equal to each other:
6x² + 5x - 13 = -2x - 10
Next, we can rearrange the equation to bring all the terms to one side:
6x² + 7x - 3 = 0
Using the quadratic formula, we can find the values of x that satisfy the equation:
x = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values a=6, b=7, and c=-3 into the formula, we get:
x = (-7 ± √(7² - 4(6)(-3))) / (2(6))
Simplifying further, we have:
x = (-7 ± √(49 + 72)) / 12
x = (-7 ± √121) / 12
x = (-7 ± 11) / 12
Therefore, the two solutions for x are:
x = (-7 + 11) / 12 = 4/12 = 1/3
x = (-7 - 11) / 12 = -18/12 = -3/2
Substituting these values of x back into either of the original equations will give us the corresponding values of y.
Learn more about Solving systems of equations algebraically