solve the equation `(x-a)/(x+a) = (x+a)/(2x-a)`  for x, when a is a constant.

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baxthum8's profile pic

baxthum8 | High School Teacher | (Level 3) Associate Educator

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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(x-5a)=0`

`x = 0 or x-5a =0`

`x = 0 or x = 5a`

``

``

givingiswinning's profile pic

givingiswinning | Student, Grade 10 | (Level 1) Valedictorian

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(x-a)(2x-a)=(x+a)(x+a)

2x^2-ax-2ax+a^2=x^2+ax+ax+a^2

2x^2-x^2-ax-2ax-ax-ax+a^2-a^2=0

x^2-5ax=0

x-5a=0

x=5a; x=0

taangerine's profile pic

taangerine | Student, Grade 12 | (Level 1) Valedictorian

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Cross Multiply 

  • `(x-a)(2x-a)=(x+a)^2`

FOIL Both sides (First, Outside, Inside, Last) 

  • `2x^(2)-xa-2xa+a^(2)=x^(2)+xa+xa+a^(2)`

Simplify

  • `2x^(2)-3xa+a^(2)=x^(2)+2xa+a^(2)`

Bring all terms to one side and simplify

  • `x^(2)-5xa=0`

Q: What do the 2 terms have in common?

A:  `x`

`x(x-5a)=0`

`x=0, 5a`

Because the 2 terms have x in common, we can factor out the x on both sides then simplify. 

mchandrea's profile pic

mchandrea | Student, College Junior | (Level 1) Salutatorian

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`(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

` `

` `

atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

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cross multiply:

(x-a)(2x-a) = (x+a)(x+a)

foil

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

factor our

x(x-5a)

set them equal to 0

x=0

x-5a=0

x= 5a

glendamaem's profile pic

glendamaem | Student | (Level 1) Honors

Posted on

cross multiply:

(x-a)(2x-a)=(x+a)(x+a)

2x^2-ax-2ax+a^2=x^2+ax+ax+a^2

bring all variables to the left side of the equation and equate to zero (transpose)

2x^2-x^2-ax-2ax-ax-ax+a^2-a^2=0

x^2-5ax=0

x-5a=0

x=5a; x=0

Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

Cross multiply to get rid of the fractions: 
(x-a)(2x - a) = (x + a)(x + a)
 
FOIL the binomials on both sides: 
 
`2x^2 -3ax + a^2 = x^2 + 2ax + a^2`
``
 Bring all the left hand terms to the right side and simplify: 
 
`x^2 - 5ax = 0`
``
``Solve for x: 
 
x ( x - 5a ) = 0
 
x = 0    and x = 5a
ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

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`(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> `

rimmery's profile pic

rimmery | Student, Undergraduate | (Level 1) Honors

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`(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`

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