Question

How do you solve `\frac { 1} { x - 7} + \frac { x + 2} { x - 3} = \frac { 3} { x ^ { 2} - 10x + 21}`?

Answer 1

`x=-2.90 or 6.90`

Explanation:

First, solve the quadratic on the bottom of the fraction (denominator) on the left side:

`3/(x^2-10x+21)`

`3/((x-3)(x-7))`

Now we move to the right side and add the two fractions together by multiplying the two denominators together.

`(x-3+(x+2)(x-7))/((x-3)(x-7))`

Now we get this:

`(x-3+(x+2)(x-7))/((x-3)(x-7))=3/((x-3)(x-7))`

You can see that the two denominators are the same so we can eliminate them to make our lives easier:

`x-3+(x+2)(x-7)=3`

Expand the brackets and simplify by bringing everything over to one side:

`x^2-4x-20=0`

Plot this on a graph (using a graphing calculator) and you will see that the graph crosses the `x`-axis at `-2.9` and `6.9`.

There we go, those are the two answers. I also checked the equation and everything seems in order.

Hope this helps,

SRNAG (some random nerd across the globe)

Answer 2

`x=2+-2sqrt6`

Explanation:

First, we can factor the denominator on the right by grouping:
`1/(x-7)+(x+2)/(x-3)=3/(x^2-10x+21)`

`1/(x-7)+(x+2)/(x-3)=3/(x^2-7x-3x+21)`

`1/(x-7)+(x+2)/(x-3)=3/(x(x-7)-3(x-7))`

`1/(x-7)+(x+2)/(x-3)=3/((x-7)(x-3))`

Next, we multiply through by `(x-7)(x-3)`:
`(x-7)(x-3)(1/(x-7)+(x+2)/(x-3))=3/cancel((x-7)(x-3))*cancel((x-7)(x-3))`

`x-3+(x+2)(x-7)=3`

`x-3+x^2-7x+2x-14=3`

`x^2-4x-20=0`

Solve with the quadratic formula:
`x=2+-2sqrt6`

Answer 3

See a solution process below:

Explanation:

First, put both fractions on the left side of the equation over a common denominator by multiplying each by the appropriate form of `1`:

`((x - 3)/(x - 3) xx 1/(x - 7)) + ((x - 7)/(x - 7) xx (x + 2)/(x - 3)) = 3/(x^2 - 10x + 21)`

`((x - 3) xx 1)/((x - 3) xx (x - 7)) + ((x - 7) xx (x + 2))/((x - 7) xx (x - 3)) = 3/(x^2 - 10x + 21)`

`(x - 3)/(x^2 - 3x - 7x + 21) + (x^2 + 2x - 7x - 14)/(x^2 - 7x - 3x + 21) = 3/(x^2 - 10x + 21)`

`(x - 3)/(x^2 - 10x + 21) + (x^2 - 5x - 14)/(x^2 - 10x + 21) = 3/(x^2 - 10x + 21)`

`((x - 3) + (x^2 - 5x - 14))/(x^2 - 10x + 21) = 3/(x^2 - 10x + 21)`

`(x - 3 + x^2 - 5x - 14)/(x^2 - 10x + 21) = 3/(x^2 - 10x + 21)`

`(x^2 - 5x + x - 14 - 3)/(x^2 - 10x + 21) = 3/(x^2 - 10x + 21)`

`(x^2 - 4x - 17)/(x^2 - 10x + 21) = 3/(x^2 - 10x + 21)`

Next, multiply each side of the equation by `color(red)((x^2 - 10x + 21))` to eliminate the fractions while keeping the equation balanced:

`color(red)((x^2 - 10x + 21)) xx (x^2 - 4x - 17)/(x^2 - 10x + 21) = color(red)((x^2 - 10x + 21)) xx 3/(x^2 - 10x + 21)`

`cancel(color(red)((x^2 - 10x + 21))) xx (x^2 - 4x - 17)/color(red)(cancel(color(black)(x^2 - 10x + 21))) = cancel(color(red)((x^2 - 10x + 21))) xx 3/color(red)(cancel(color(black)(x^2 - 10x + 21)))`

`x^2 - 4x - 17 = 3`

Then, subtract `color(red)(3)` from each side of the equation to put the equation in standard form while keeping the equation balanced:

`x^2 - 4x - 17 - color(red)(3) = 3 - color(red)(3)`

`x^2 - 4x - 20 = 0`

Now, we can use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(1)` for `color(red)(a)`

`color(blue)(-4)` for `color(blue)(b)`

`color(green)(-20)` for `color(green)(c)` gives:

`x = (-color(blue)(-4) +- sqrt(color(blue)(-4)^2 - (4 * color(red)(1) * color(green)(-20))))/(2 * color(red)(1))`

`x = (4 +- sqrt(16 - (-80)))/2`

`x = 4/2 +- sqrt(16 + 80)/2`

`x = 2 +- sqrt(96)/2`

`x = 2 +- sqrt(16 * 6)/2`

`x = 2 +- (sqrt(16)sqrt(6))/2`

`x = 2 +- (4sqrt(6))/2`

`x = 2 +- 2sqrt(6)`