Use the quadratic formula to find the real solutions:
2x/x-3 + 2/x = 3
I know I multiply both sides by x(x-3), but I don't know how to do that.
rewrite the equation as ...
[2x/(x-3)](x(x-3)) + [2/x](x(x-3)) = 3(x(x-3))
notice in the first term, x-3 on top cancels x-3 at the bottom, in the second term x on top cancels x at the bottom , to get ...
2x(x) + 2(x-3) = 3x^2 - 9x
2x^2 + 2x - 6 = 3x^2 - 9x
x^2 - 11x + 6 = 0
x = (11 ± √97)/2
thank you!
To solve the given equation, you need to first multiply both sides of the equation by the common denominator, which is x(x-3). Let's go through the steps:
1. Start with the equation: 2x/(x-3) + 2/x = 3.
2. Multiply both sides of the equation by the common denominator, x(x-3), to eliminate the denominators. The equation becomes:
x(x-3) * (2x/(x-3)) + x(x-3) * (2/x) = x(x-3) * 3.
3. Simplify each term on both sides:
2x^2 + 2(x)(x-3) = 3x(x-3).
4. Expand and collect like terms:
2x^2 + 2x(x-3) = 3x^2 - 9x.
2x^2 + 2x^2 - 6x = 3x^2 - 9x.
4x^2 - 6x = 3x^2 - 9x.
5. Move all the terms to one side of the equation, so that the equation becomes quadratic:
4x^2 - 6x - 3x^2 + 9x = 0.
x^2 + 3x = 0.
6. Now, we have a quadratic equation of the form ax^2 + bx = 0, where a = 1 and b = 3. Apply the quadratic formula to find the values of x:
x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, substitute a = 1, b = 3, and c = 0:
x = (-3 ± √(3^2 - 4(1)(0))) / (2(1)).
x = (-3 ± √(9)) / (2).
x = (-3 ± 3) / 2.
7. Simplify each root:
x1 = (-3 + 3) / 2 = 0 / 2 = 0.
x2 = (-3 - 3) / 2 = -6 / 2 = -3.
So, the real solutions to the quadratic equation 2x/(x-3) + 2/x = 3 are x = 0 and x = -3.