Solve .

First find the LCM of the denominators → (x + 1)(x + 3).

Now multiply everything by the LCM.

Next, cancel like factors.

2(x + 3) – 3(x + 1) = -4(x + 3)

Multiply it out.

2x + 6 – 3x – 3 = -4x – 12

Collect like terms.

3 – x = -4x – 12

3x = -15

x = -5

Check: Does -5 make any denominators zero?

No, so x = -5 is a real solution.

Solve .

The LCM of the denominators is (x + 3)(x – 3), so let's multiply that by every stinkin' term.

Now cancel what you can.

x + 3 – (x – 3) = x + 3

x + 3 – x + 3 = x + 3

6 = x + 3

x = 3

But is it an extraneous solution or a real solution?

Well, does it make any of the denominators zero?

Yes, 3 – 3 = 0. So yeah, x = 3 is an extraneous solution.

The answer is that there is no solution.

Solve .

What's the LCM of the denominators?

(x – 2)(x – 4)

Now multiply through with it.

Cancel…

8 – 2(x – 4) = x – 2

8 – 2x + 8 = x – 2

16 – 2x = x – 2

18 = 3x

x = 6

Is it real or extraneous?

Does is make any of the denominators zero?

No. It's all good: x = 6 is a valid solution.

Solve .

We can factor that first denominator to make our lives easier:

x² – 25 = (x – 5)(x + 5)

That means our LCM is (x – 5)(x + 5), so that's what we'll multiply by.

Cancel like a boss.

10 = 5(x + 5)

10 = 5x + 25

5x = -15

x = -3

Is it a real solution?

Yep, -3 does not make either of the denominators zero.

The answer is x = -3.

Solve .

Multiply through by the LCM of x(x – 1), then cancel like your life depends on it. That leaves us with:

2x – x(x – 1) = 2(x – 1)

Multiply things out.

2x – x² + x = 2x – 2

Collect like terms.

3x – x²= 2x – 2

0 = x² – x – 2

x² – x – 2 = 0

Now factor that quadratic.

(x – 2)(x + 1) = 0

Set each factor to zero and solve:

x – 2 = 0 and x + 1 = 0

x = 2 and x = -1

Are they real solutions or extraneous solutions?

Neither of 'em makes any denominator zero, so both are real.