college algebra can't find answer

[kfrohnapfel]

i can't seem to find the answer to one problem on my homework. help please!
steps included with the answer would be helpful as well

x divided by (x-3) plus (x^2-2) divided by (9-x^2) = 1 divided by (x+3)

 

[BigGlenntheHeavy]

\(\displaystyle \frac{x}{x-3}+\frac{x^2-2}{9-x^2} \ = \ \frac{1}{x+3}\)

\(\displaystyle = \ \frac{x}{x-3}-\frac{x^2-2}{x^2-9} \ = \ \frac{1}{x+3}\)

\(\displaystyle = \ \frac{x}{x-3}-\frac{x^2-2}{(x+3)(x-3)} \ = \ \frac{1}{x+3}\)

\(\displaystyle =x(x+3)+2-x^2 \ = \ x-3\)

\(\displaystyle = \ x^2+3x+2-x^2 \ = \ x-3\)

\(\displaystyle = \ 2x \ = \ -5, \ \implies \ x \ = \ -\frac{5}{2}\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ you.\)

 

[mmm4444bot]

Here is one possible sequence of steps, to get you to a simple linear equation.

Step 1: Recognize that the denominator in the ratio (x^2 - 2)/(9 - x^2) factors as a difference of squares.

Step 2: Factor 9 - x^2 to rewrite the ratio as (x^2 - 2)/[(3 - x)(3 + x)]

Step 3: Recognize that (x + 3)(x - 3) would be the LCD, if we could somehow change (3 - x) to (x - 3)

Step 4: Realize that (-1)(3 - x) is (x - 3)

Step 5: Multiply the ratio by (-1)/(-1) to obtain -(x^2 - 2)/[(x + 3)(x - 3)]

The equation is now:

\(\displaystyle \frac{1}{x + 3} \;=\; \frac{x}{x - 3} \;-\; \frac{x^2 - 2}{(x + 3)(x - 3)}\)

Step 6: Multiply both sides by the LCD

Step 7: Solve the resulting linear equation for x.

If you would like more help, please show how far you're able to get, so that I know where to continue helping.

 

[kfrohnapfel]

thanks

 

[Denis]

x divided by (x-3) plus (x^2-2) divided by (9-x^2) = 1 divided by (x+3)

Another way; your equation:
x / (x-3) + (x^2-2)/(9-x^2) = 1 / (x+3)
Rearrange:
x / (x-3) - 1 / (x+3) = (x^2 - 2) / (x^2-9)

[x(x+3) - (x-3)] / [(x-3)(x+3)] = (x^2 - 2) / [(x-3)(x+3)]

x(x+3) - (x-3) = x^2 - 2

x^2 + 3x - x + 3 = x^2 - 2 ; finish it