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2x-a/x+a - x+3a/a-x=a^2/a^2-x^2 + 3
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# 2x-a/x+a - x+3a/a-x=a^2/a^2-x^2 + 3

[From: ] [author: ] [Date: 11-05-02] [Hit: ]
C.D.[(2x-a)(a-x)-(x-3a)(x+a)]/(a²-x²)=a²/(…+3 [-2x²+3ax-a²-(x²-2ax-3a²)]=a²+3(a²-x²)-3x²+2a²+5ax=a²+3a²-3x²5ax=2a² x=2/5 a-fully factor everything.SINCE denominators are equal to each other, solve using numerators.use quadratic formula.......
Solve
A.) X= -8a
B.) X=-4a
C.) X=4a
D.) X=8a

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(2x-a)/(x+a) -(x+3a)/(a-x)=a²/(a²-x²) +3
[(2x-a)(a-x)-(x-3a)(x+a)]/(a²-x²)=a²/(… +3
[-2x²+3ax-a²-(x²-2ax-3a²)]=a²+3(a²-x²)
-3x²+2a²+5ax=a²+3a²-3x²
5ax=2a²
x=2/5 a

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fully factor everything.
assuming expression is

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

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

lowest comon denominator: (x-a)(x+a):

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

SINCE denominators are equal to each other, solve using numerators.

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

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

2x^2 +ax -a^2 -(3a^2 +4ax +x^2) = a^2 +3(a^2 -x^2)

2x^2 +ax - a^2 -3a^2 -4ax -x^2 = a^2 +3a^2 -3x^2

x^2 -4a^2 -3ax = 4a^2 -3x^2

0 = 4a^2 +4a^2 -3x^2 -x^2 +3ax

0 = 8a^2 +3ax - 4x^2