solve the equation `(x-a)/(x+a) = (x+a)/(2x-a)` for x, when a is a constant.
Cross multiply to get: ``
`(x-a)(2x-a) = (x+a)^2`
`2x^2-3ax+a^2 = x^2+2ax + a^2`
`x^2 -5ax = 0`
`x = 0 or x-5a =0`
`x = 0 or x = 5a`
FOIL Both sides (First, Outside, Inside, Last)
Bring all terms to one side and simplify
Q: What do the 2 terms have in common?
Because the 2 terms have x in common, we can factor out the x on both sides then simplify.
`(x-a)/(x+a) = (x+a)/(2x-a)` the firs step is to cross multiply
`2x^2 - 3xa + a^2 = x^2 + 2xa +a^2` expand the answers
`2x^2 - 3xa + a^2 - x^2 - 2xa - a^2` = 0 transpose all the factors to one side
`x^2 - 5xa = 0` add or subtract all possible factors and equate with 0
`x (x - 5a) = 0` get the common factor
`x = 0 x=5a` get the value of x
The final answer should be x = 0 and x = 5a
(x-a)(2x-a) = (x+a)(x+a)
2x^2 - ax - 2ax +a^2 = x^2 +ax +ax+a^2
combine like terms:
2x^2 -3ax + a^2 = x^2 + 2ax + a^2
move them all to one side:
2x^2 -3ax + a^2 - x^2 - 2ax - a^2
combine like terms:
x^2 - 5ax
set them equal to 0
bring all variables to the left side of the equation and equate to zero (transpose)
`(x-a)/(x+a)` = `(x+a)/(2x-a)` Your first step is to cross multiply, which is taking the denominator and multiplying it to the equation that is on the opposite of the equal sign.
`(x-a)(2x-a)` = `(x+a)(x+a)` This is what it looks like after cross-multiplying.
The next step is to FOIL (first, outer, inner, last) the equation. Remember the signs of the equations and what happens when you multiply two negatives!
`2x^2 - ax -2xa +a^2` = `x^2 + 2ax + a^2` You can see that this is not in the simplest form. To make this easier, combine like terms such as the `ax` .
`2x^2 - 3ax + a^2` = You should also notice that both sides of the equation has `a^2` . When something is on both sides of the equation, you can simply ignore it.
`2x^2 - 3ax` = `x^2 + 2ax` Set the entire equation to zero by moving one side to the other. I prefer keeping the largest polynomial a positive number.
`x^2 - 5ax` = `0` You can factor out the `x` and set equal to zero.
`(x)(x-5a)` = `0`
`x = 0` and `x-5a = 0` Finally, solve for `x` .
Your answer should be: `x=0` and `x=5a` .` <br> `
`(x-a)(2x-a)=(x+a)^2` (Cross multiply)
`2x^2-ax-2ax+a^2=x^2+2ax+a^2` (Expand both sides)
`2x^2-3ax+a^2=x^2+2ax+a^2` (Simplify equation)
`x^2-5ax=0` (Simplify equation)
`x(x-5a)=0` (Factor out the x from left-hand side)
From the equation above, we see that either:
`x=0 or x-5a=0`
When `x-5a=0`, we get that `x=5a`
So the final answer is `x=0 or x=5a`