To solve this system of equations, we can substitute one equation into the other to find the value of one of the variables. Let's substitute the second equation into the first:
y = -5x + 4a^2x^2 + 34a - 66 = -5x + 4a^2
Rearrange the equation to isolate x:
x^2 + 5x + 34a - 4a^2 - 66 = 0x^2 + 5x - 4a^2 + 34a - 66 = 0
Now we have a quadratic equation in terms of x. To solve for x, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
With a = 1, b = 5, and c = -4a^2 + 34a - 66, we can plug these values into the formula to find the solutions for x.
To solve this system of equations, we can substitute one equation into the other to find the value of one of the variables. Let's substitute the second equation into the first:
y = -5x + 4a^2
x^2 + 34a - 66 = -5x + 4a^2
Rearrange the equation to isolate x:
x^2 + 5x + 34a - 4a^2 - 66 = 0
x^2 + 5x - 4a^2 + 34a - 66 = 0
Now we have a quadratic equation in terms of x. To solve for x, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
With a = 1, b = 5, and c = -4a^2 + 34a - 66, we can plug these values into the formula to find the solutions for x.