Question

How do you solve `\frac { e } { e + 5} + \frac { 5} { e - 5} = \frac { 2e + 24} { e ^ { 2} - 25}`?

Answer 1

`e=1`

Explanation:

difference of two squares identity:

`(a+b)(a-b) = a^2-b^2`

using this:

`(e+5)(e-5) = e^2-25`

this can be used as a common denominator:

`e/(e+5) = (e(e-5))/((e+5)(e-5)`

`=(e(e-5))/(e^2-25)`

`5/(e-5) = (5(e+5))/(e^2-25)`

`=(e(e-5))/(e^2-25)+ (5(e+5))/(e^2-25) = (2e+24)/(e^2-25)`

multiply everything by `e^2-25`:

`e(e-5)+5(e+5)=2e+24`

expand brackets:

`e^2cancel(-5e)cancel(+5e)+25 = 2e+24`

`e^2+25 = 2e+24`

subtract `24`:

`e^2+1=2e`

subtract `2e`:

`e^2-2e+1 = 0`

`-1 + -1 = -2`
`-1 * -1 = 1`

`(e-1)^2 = 0`

`e=1`

graphical proof:

graph{(x-1)^2 [-0.434, 2.233, -0.479, 0.855]}

equation of graph: `y=(x-1)^2`

vertex of graph: `(1,0)`
`(x=1, y=0)`

substitute `x` with `e`
`e=1`

Answer 2

`e=1`

Explanation:

In equations with fractions, you can get rid of the denominators right at the beginning by multiplying each term by the LCD.

Factorise the denominators if possible.

`e/((e+5))+5/((e-5)) = (2e+24)/((e+5)(e-5)`

`LCD = color(blue)((e+5)(e-5))`

Look at the left hand side first:
`(color(blue)(cancel((e+5))(e-5)))/cancel((e+5))+(5color(blue)((e+5)cancel((e-5))))/(cancel((e-5)))`

Look at the right hand side

`= ((2e+24)color(blue)cancel((e+5)(e-5)))/cancel((e+5)(e-5)`#

After cancelling the denominators we have:

`e-5 +5(e+5) = 2e+24`

`e-5 +5e+25 = 2e+24`

`6e-2e= 24+5-25`

`4e = 4`

`e=1`