Question

How do you evaluate `\frac { x - 3} { 2x - 10} - \frac { 9x + 57} { 10x + 50} = \frac { x - 7} { x ^ { 2} - 25}`?

Answer 1

Solutions are `x=-11"; "x=+8`

`x=5` is an excluded value

Explanation:

Page width causes equations to take up more than 1 line

I factored out the numerators. However, later on it turns out that factoring the numerator does not help. Left in place rather than change everything.

`" "(x-3)/(2(x-5))" " -" "(3(3x+19))/(10(x+5))" "=(x-7)/((x-5)(x+5))`

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Building common denominators:

`color(green)([(x-3)/(2(x-5))color(red)(xx1)] -[(3(3x+19))/(10(x+5))color(red)(xx1)]=[(x-7)/((x-5)(x+5))color(red)(xx1)])`

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`color(green)([(x-3)/(2(x-5))color(red)(xx(5(x+1))/(5(x+1)))] -[(3(3x+19))/(10(x+5))color(red)(xx(x-5)/(x-5))]=[(x-7)/((x-5)(x+5))color(red)(xx10/10)])`
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`color(green)([(5(x+1)(x-3))/(10(x-5)(x+5))] -[(3(3x+19)(x-5))/(10(x-5)(x+5))]=[(10(x-7))/(10(x-5)(x+5))])`

Note that at `x=5` the denominator part `(x-5)=0`. Thus the wole denominator becomes 0. Consequently the equation is 'undefined' at `x=5`
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Multiply both sides by `10(x-5)(x+5)` cancelling out the denominator.

`[5(x+1)(x-3)] - [3(3x+19)(x-5)] = 10(x-7)`

`[5x^2-14x-3]-[9x^2+12x-285]=10x-70`

`-4x^2 -2x +282=10x-70`

`-4x^2-12x+352=0`

`-x^2-3x+88=0`

Notice that `11xx8=88" and "11-8=3`

The `x^2` is negative so end up with `-3x` we have:

`(x+11)(-x+8)=0`

Solutions are `x=-11"; "x=+8`