Solve for 𝑥 and 𝑦 simultaneously: 2𝑦 − 𝑥 = 4 and − 𝑥2 − 𝑥 + 12 = 𝑦

Answer 1

To solve for 𝑥 and 𝑦 simultaneously, we will substitute the second equation into the first equation.

From the second equation, we have 𝑦 = -𝑥^2 - 𝑥 + 12.

Now substitute 𝑦 = -𝑥^2 - 𝑥 + 12 into the first equation:

2(-𝑥^2 - 𝑥 + 12) - 𝑥 = 4
-2𝑥^2 - 2𝑥 + 24 - 𝑥 = 4
-2𝑥^2 - 3𝑥 + 20 = 0

Now let's solve this quadratic equation using the quadratic formula:

𝑥 = (-(-3) ± √((-3)^2 - 4*(-2)*20)) / 2*(-2)
𝑥 = (3 ± √(9 + 160)) / -4
𝑥 = (3 ± √169) / -4
𝑥 = (3 ± 13) / -4

Therefore, we have two potential solutions for 𝑥:
𝑥 = (3 + 13) / -4 = -16 / -4 = 4
𝑥 = (3 - 13) / -4 = -10 / -4 = 2.5

Now that we have the values for 𝑥, we can substitute them back into the second equation to solve for 𝑦:

For 𝑥 = 4:
𝑦 = -(4)^2 - 4 + 12
𝑦 = -16 - 4 + 12
𝑦 = -8

For 𝑥 = 2.5:
𝑦 = -(2.5)^2 - 2.5 + 12
𝑦 = -6.25 - 2.5 + 12
𝑦 = 3.25

Therefore, the solutions to the system of equations are:
𝑥 = 4, 𝑦 = -8
𝑥 = 2.5, 𝑦 = 3.25