How do you solve $\frac{x + 3}{x^{2} + 3 x - 4} = \frac{x + 2}{x^{2} - 16}$ $(x+3)/(x^(2)+3x-4)=(x+2)/(x^(2)-16)$ ?

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Answer 1 · 2015-05-31T14:47:13

$\frac{x + 3}{x^{2} + 3 x - 4} = \frac{x + 2}{x^{2} - 16}$ $(x+3)/(x^(2)+3x-4)=(x+2)/(x^(2)-16)$

By property: $a^{2} - b^{2} = (a + b) (a - b)$ $color(blue)(a^2 - b^2 = (a+b)(a-b)$

so, $x^{2} - 16 = (x + 4) (x - 4)$ $x^(2)-16 = color(blue)((x+4)(x-4)$

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

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

How do you solve (x+3)/(x^(2)+3x-4)=(x+2)/(x^(2)-16)x+3x2+3x−4=x+2x2−16(x+3)/(x^(2)+3x-4)=(x+2)/(x^(2)-16)?