Solve the system: x = y2 + 4y – 5, x = 0.
Question
Solve the system: x = y2 + 4y – 5, x = 0.
Solution
Step 1: We are given two equations, x = y^2 + 4y - 5 and x = 0.
Step 2: Since x = 0, we can substitute x in the first equation with 0.
So, 0 = y^2 + 4y - 5.
Step 3: Now, we need to solve this quadratic equation for y.
Step 4: To solve the quadratic equation, we can use the quadratic formula, y = [-b ± sqrt(b^2 - 4ac)] / 2a.
In this equation, a = 1, b = 4, and c = -5.
Step 5: Substituting these values into the quadratic formula gives us:
y = [-4 ± sqrt((4)^2 - 41(-5))] / 2*1 y = [-4 ± sqrt(16 + 20)] / 2 y = [-4 ± sqrt(36)] / 2 y = [-4 ± 6] / 2
Step 6: This gives us two solutions:
y = ( -4 + 6 ) / 2 = 1 y = ( -4 - 6 ) / 2 = -5
So, the solutions to the system of equations are x = 0, y = 1 and x = 0, y = -5.
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