Question

How do you solve `(3x+2)/(3x-2)=(4x-7)/(4x+7)`?

Answer 1

x = 0

Explanation:

Awesome fact :

If and only if `(a+b)/(a-b) = (c+d)/(c-d)`

then

`ad = bc`

if you have habits with math, you know the only solution possible is : `0`

you have

`(3x+2)/(3x-2)=(4x-7)/(4x+7)` `=>`

`(3x+2)/(3x-2)=(-4x+7)/(-4x-7)`

`u = -4x`
`v = 3x`

`(v+2)/(v-2)=(u+7)/(u-7)`

then :

`7v = 2u`

`21x = -8x`

`29x = 0`

`x = 0`

Answer 2

.
`color(green)(x=0)` ..A different approach. Really this is the same thing as Tom's. The difference is that he has jumped steps and simplified using substitution.

Explanation:

Given: `color(brown)( (3x+2)/(3x-2) = (4x-7)/(4x+7)`

`color(blue)("'Getting rid' of the denominators")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(green)("Multiply both sides by "color(blue)(3x-2))`
`color(brown)( (3x+2)/(3x-2)color(blue)(xx(3x-2))= (4x-7)/(4x+7)color(blue)(xx(3x-2))`

`color(brown)( (3x+2)xx color(blue)((3x-2))/((3x-2))=((4x-7)color(blue)((3x-2)))/(4x+7))`

but `(3x-2)/(3x-2)` is another way of writing 1 giving:

`(3x+2) xx 1= ((4x-7)(3x-2))/(4x+7)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(green)("Multiply both sides by "color(blue)(4x+7))`

`color(brown)((3x+2) color(blue)(xx(4x+7))= ((4x-7)(3x-2))/(4x+7)color(blue)(xx(4x+7))`

`(3x+2)(4x+7)=(4x-7)(3x-2)xx((4x+7))/((4x+7))`

But`(4x+7)/(4x+7)` is another way of writing 1 giving:

`color(brown)((3x+2)color(blue)((4x+7))=(4x-7)color(black)((3x-2)))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(green)("Multiply out the brackets")`
`color(brown)(3xcolor(blue)((4x+7))+2color(blue)((4x+7)) =4x color(black)((3x-2))-7color(black)((3x-2))`

`12x^2+21x+8x+14=12x^2-8x-21x+14`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(green)("Collecting like terms")`

`(12x^2-12x^2)+(21x+8x+8x+21x)=14-14`

`0x^2 +58x=0`

`color(green)(x=0)`