SOLUTION: {{{x^2/x-2=4/x-2}}} I don't know if I typed this question out right. It should be x squared over x-2 = 4 over x-2. And the instructions are to solve for x. Thank you shelena

Question 126848: I don't know if I typed this question out right. It should be x squared over x-2 = 4 over x-2. And the instructions are to solve for x.
Thank you
shelena
Found 2 solutions by MathLover1, bucky:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

....if denominators are same, then nominators must be too if we have equal sign, so you have:

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given:
.

.
Since the denominators of both terms are we can get rid of them by multiplying
both sides of this equation by the quantity . That multiplication leads to:
.

.
Then cancel the terms in the numerators with the same terms in the denominators:
.

.
What you are left with is:
.

.
To solve, take the square root of both sides to get two possible answers:
.
and
.
because if either 2 or -2 is squared, the result is +4.
.
We need to check out these two answers. Let's look first at x = +2. Go to the original
problem and substitute +2 for x. oops. Notice what happens. Both terms have (x - 2) in the
denominator. When x is +2 and you subtract 2 from it, the denominator becomes 0. But division
by zero is not allowed in algebra. Therefore, x cannot equal +2.
.
What about if x = -2. If we substitute -2 for x in the original problem, it becomes:
.

.
(-2 - 2) is equal to -4. So we can substitute -4 for both of the denominators to get:
.

.
Then you can note that squaring -2 results in +4. You can substitute +4 for the numerator on
the left side. When you do that substitution the equation becomes:
.

.
Without going further, you can see that the left side equals the right side. Therefore,
if you take -2 as the value of x, it will maintain the equality of both sides of the equation.
.
Therefore, the answer to your problem is x = -2.
.
Hope this helps you to understand the problem and to see the procedures that you can use to
solve it.
.
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