SOLUTION: These two problems are causing me trouble {{{5/(x-2)- 5/(x+2)=4}}} also, {{{(x/3)-(x+2)/2 =1}}}

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: These two problems are causing me trouble {{{5/(x-2)- 5/(x+2)=4}}} also, {{{(x/3)-(x+2)/2 =1}}}      Log On


   



Question 110020: These two problems are causing me trouble 5%2F%28x-2%29-+5%2F%28x%2B2%29=4 also, %28x%2F3%29-%28x%2B2%29%2F2+=1
Found 2 solutions by MathLover1, edjones:
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!
1. 5%2F%28x-2%29-+5%2F%28x%2B2%29=4.......... on left side , common denominator is %28x-2%29%28x%2B2%29
%285%28x%2B2%29-+5%28x-2%29%29%2F%28x-2%29%28x%2B2%29=4..........multiply both sides by %28x-2%29%28x%2B2%29
..........
%285%28x%2B2%29-+5%28x-2%29%29+=4%28x-2%29%28x%2B2%29+..........
5%28%28x%2B2%29-+%28x-2%29%29+=4%28x-2%29%28x%2B2%29+..........
5%28x%2B2+-+x+%2B+2%29+=4%28x-2%29%28x%2B2%29+..........
5%282%2B+2%29+=4%28x-2%29%28x%2B2%29+..........
20+=4%28x-2%29%28x%2B2%29+..........divide both sides by 4
5+=%28x-2%29%28x%2B2%29+..........

5+=x%5E2-2%5E2+..........
5+=x%5E2-4+..........move -4 to the left
5+%2B+4+=x%5E2+..........
or
x%5E2+=+9
x%5B1%5D=+%2B+3
x%5B2%5D=+-+3

2. %28x%2F3%29-%28x%2B2%29%2F2+=1........common denominator is 3%2A2=6

%282x-3%2A%28x%2B2%29%29%2F6+=1........multiply both sides by 6
6%2A%282x-3%2A%28x%2B2%29%29%2F6+=1%2A6........
%282x-3%2A%28x%2B2%29%29+=+6........
%282x-3x+-6%29+=+6........
-x+-6+=6........move -6 to the right
-x+=6+%2B+6........multiply both sides by -1
x+=+-12........


Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
In all of the rational expressions get rid of fractions! makes problem much easier.
5%2F%28x-2%29-+5%2F%28x%2B2%29=4
5(x+2)-5(x-2)=4(x-2)(x+2) multiply both sides by (x-2)(x+2)
5x+10-5x+10=4(x^2-4)
+20=4(x^2-4)
x^2-4=5 divide both sides by 4
x^2=9
x=+-3
.
This is much easier.
%28x%2F3%29-%28x%2B2%29%2F2+=1
2x-3(x+2)=6 multiply each side by 6
2x-3x-6=6
-x=12
x=-12
.
Ed