Final answer:
To solve the given system of equations algebraically, we can substitute one equation into the other to eliminate one variable. By rearranging and factoring, or by using the quadratic formula, we can solve for x. Once we find the value(s) of x, we can substitute it back into either equation to find the corresponding values of y.
Step-by-step explanation:
To solve the given system of equations algebraically, we can substitute one equation into the other to eliminate one variable.
From the equation y = x², we can substitute this value for y in the second equation:
x² = -3x + 5
Now we have a quadratic equation. By rearranging and factoring, or by using the quadratic formula, we can solve for x. Once we find the value(s) of x, we can substitute it back into either equation to find the corresponding values of y.
Let's solve the quadratic equation:
x² + 3x - 5 = 0
Using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a), where a = 1, b = 3, and c = -5.
Substituting these values into the formula, we get x = (-3 ± √(9 + 20)) / 2. Simplifying this further, we have the two potential values for x: x = (-3 ± √29) / 2.
Now, substituting these values into either equation, we can find the corresponding values for y. For example, when x = (-3 + √29) / 2:
y = x²
y = [(-3 + √29) / 2]²
y = [(-3 + √29) / 2] * [(-3 + √29) / 2]
y = [(-3)² + 2(-3)(√29) + (√29)²] / 4
y = [9 - 6√29 + 29] / 4
y = [38 - 6√29] / 4
y = 19/2 - (3/2)√29
Therefore, one of the solutions to the system of equations is x = (-3 + √29) / 2 and y = 19/2 - (3/2)√29. Similarly, we can find the other solution by substituting the other value of x into either equation.
Learn more about Solving systems of equations algebraically