How do you solve for x in # 7/(x-2) + 3/(x+2) = -x/(x^2-4 )#?
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Here is your answer #x= -8 / 11 #
#(7(x+2) + 3 (x-2))/((x-2)(x+2))=-x/(x^2-4)#
#(7x+14+3x-6)/(x^2+2x-2x-4)=-x/(x^2-4)#
#(10x+8)/cancel(x^2-4)=-x/cancel(x^2-4)#
#10x + 8 = -x #
# 11 x = -8#
#x= -8 / 11 #
Given equation:
#\frac{7}{x-2}+3/{x+2}=-x/{x^2-4}#
#\frac{7(x+2)+3(x-2)}{(x-2)(x+2)}=-x/{x^2-4}#
#\frac{10x+8}{x^2-4}=-x/{x^2-4}#
#\frac{10x+8}{x^2-4}+x/{x^2-4}=0#
#\frac{10x+8+x}{x^2-4}=0#
#\frac{11x+8}{x^2-4}=0#
#11x+8=0\ \quad(\forall \ \ x\ne\pm2 )#
#11x=-8#
#x=-8/11#
It must be #x\ne 2,x\ne-2#
multimplying by #(x-2)(x+2)# we get
#7(x+2)+3(x-2)=-x# expanding we get
#7x+14+3x-6=-x#
combining like Terms
#10x+8=-x#
adding #-8# and #x#
We get
#11x=-8#
so #x=-8/11#
# 7/(x-2) + 3/(x+2) = -x/color(blue)((x^2-4 ))" "larr# factorise
# 7/(x-2) + 3/(x+2) = -x/color(blue)((x+2)(x-2))#
Multiply each tern in the equation by the LCM of the denominators which is #color(magenta)((x+2)(x-2))# and cancel where possible:
#(color(magenta)((x+2)cancel((x-2)))xx7)/cancel((x-2)) + (3xxcolor(magenta)(cancel((x+2))(x-2)))/cancel((x+2)) = (-x xxcolor(magenta)cancel((x+2)(x-2)))/cancel((x+2)(x-2))#
This leaves:
#7(x+2)+3(x-2) =-x#
#7x+14 +3x-6 =-x#
#10x+x = -8#
#11x=-8#
#x =- 8/11#
