Solve: xy=12
y=2x+2
Are the solutions x= -3, x=2, y= -4, y= 6?
Yes, though I would order them more clearly:
x= -3, y= -4
and
x=2, y= 6
Thank you
You're welcome!
To solve the system of equations:
1. Start with the second equation: y = 2x + 2.
2. Substitute this value of y into the first equation: xy = 12.
3. Replace y with 2x + 2 in the first equation: x(2x + 2) = 12.
4. Expand the equation: 2x^2 + 2x = 12.
5. Rewrite the equation in standard form: 2x^2 + 2x - 12 = 0.
6. Factor the quadratic equation: 2(x^2 + x - 6) = 0.
7. Solve for x by factoring or using the quadratic formula.
Factoring method: (x + 3)(x - 2) = 0.
x + 3 = 0 -> x = -3
x - 2 = 0 -> x = 2
Quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
a = 1, b = 1, c = -6
x = (-1 ± √(1^2 - 4(1)(-6))) / (2(1))
x = (-1 ± √(1 + 24)) / 2
x = (-1 ± √25) / 2
x = (-1 ± 5) / 2
x = -3/2 or x = 2
8. Now that we have the values of x, substitute them back into the second equation to find the corresponding values of y.
For x = -3:
y = 2(-3) + 2
y = -6 + 2
y = -4
For x = 2:
y = 2(2) + 2
y = 4 + 2
y = 6
9. The solutions to the system of equations are:
x = -3, y = -4
x = 2, y = 6
Therefore, the given solutions x = -3, x = 2, y = -4, y = 6 are correct.