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

3 Answers
Jan 22, 2018

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)

Jan 22, 2018

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

Jan 22, 2018

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)