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If a solution gives us a fraction with a denominator of 0, then that solution is extraneous

x / (x + 2) = 5 / (x - 4)

Cross-multiply

x * (x - 4) = 5 * (x + 2)

x^2 - 4x = 5x + 10

x^2 - 9x - 10 = 0

x = (9 +/- sqrt(81 + 40)) / 2

x = (9 +/- sqrt(121)) / 2

x = (9 +/- 11) / 2

x = 20/2 , -2/2

x = 10 , -1

x = 10 and x = -1 are the solutions to our problem.  Let's find possible exceptions by letting the denominators of each fraction equal 0

x + 2 = 0

x = -2

x - 4 = 0

x = 4

Extraneous solutions would be x = -2  and x = 4

Since x = -2 and x = 4 are not part of our solution set, then there are no extraneous solutions

x = -1 , 10

Test them in the original problem

x / (x + 2) = 5 / (x - 4)

-1 / (-1 + 2) = 5 / (-1 - 4)

-1 / (1) = 5 / (-5)

-1 = -1

10 / (10 + 2) = 5 / (10 - 4)

10 / 12 = 5 / 6

5/6 = 5/6

Both solutions work out, so there are no extraneous solutions at all.

Solve the equation x/(x + 2) = 5/(x - 4)

x(x - 4) = 5(x + 2)

x^2 - 9x - 10 = 0

(x + 1)(x - 10) = 0

Solutions:

a) There are no extraneous solutions  

c) x = 10

d) x = -1 

Simply substitute the answers into the equation and check

4 / (4 + 2) = 5 / (4 - 4).  Does the LHS equal the RHS?  If not, it's not an answer.

Check the next and so on.

Your equation:

x / (x + 2) = 5 / (x - 4)

First, we see that x cannot be -2 or 4 to prevent dividing by zero.

Next, multiply both sides by the LCD to get rid of the fractions:

x(x - 4) = 5(x + 2)

x² - 4x = 5x + 10

x² - 9x - 10 = 0

(x - 10)(x + 1) = 0

x = -1 and 10

These aren't in our "divide by zero" checks, so we have two values for x so far. Test them against the original equation to see if any are extraneous:

x / (x + 2) = 5 / (x - 4)

-1 / (-1 + 2) = 5 / (-1 - 4) and 10 / (10 + 2) = 5 / (10 - 4)

-1 / 1 = 5 / (-5) and 10 / 12 = 5 / 6

-1 = -1 and 5 / 6 = 5 / 6

TRUE and TRUE

Both solutions check out so there are no extraneous solutions.

-2 and 4 would only count as extraneous if the solution to the quadratic contained these values.