How do you solve #(x+3)/(x^(2)+3x-4)=(x+2)/(x^(2)-16)#?

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Answer 1 · 2015-05-31T14:47:13

#(x+3)/(x^(2)+3x-4)=(x+2)/(x^(2)-16)#

By property: #color(blue)(a^2 - b^2 = (a+b)(a-b)#

so, #x^(2)-16 = color(blue)((x+4)(x-4)#

By splitting the middle term,
#x^2+3x-4# can be factorised as #color(green)((x+4)(x-1)#

now rewriting the expression
#(x+3)/color(green)((cancelx+4)(x-1))=(x+2)/color(blue)(cancel(x+4)(x-4)#

we get,
# (x+3)/(x-1) = (x+2)/(x-4)#

cross multiplying:
# (x+3).(x-4) = (x+2).(x-1)#
#cancel x^2 -x -12 = cancelx^2 +x - 2#
#- 10 = 2x , color(red)(x =-5#