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

Jan 10, 2016

x = 0

#### Explanation:

Awesome fact :

If and only if $\frac{a + b}{a - b} = \frac{c + d}{c - d}$

then

$a d = b c$

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

you have

$\frac{3 x + 2}{3 x - 2} = \frac{4 x - 7}{4 x + 7}$ $\implies$

$\frac{3 x + 2}{3 x - 2} = \frac{- 4 x + 7}{- 4 x - 7}$

$u = - 4 x$
$v = 3 x$

$\frac{v + 2}{v - 2} = \frac{u + 7}{u - 7}$

then :

$7 v = 2 u$

$21 x = - 8 x$

$29 x = 0$

$x = 0$

Jan 13, 2016

.
$\textcolor{g r e e n}{x = 0}$ ..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)

$\textcolor{b l u e}{\text{'Getting rid' of the denominators}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Multiply both sides by } \textcolor{b l u e}{3 x - 2}}$
color(brown)( (3x+2)/(3x-2)color(blue)(xx(3x-2))= (4x-7)/(4x+7)color(blue)(xx(3x-2))

$\textcolor{b r o w n}{\left(3 x + 2\right) \times \frac{\textcolor{b l u e}{\left(3 x - 2\right)}}{\left(3 x - 2\right)} = \frac{\left(4 x - 7\right) \textcolor{b l u e}{\left(3 x - 2\right)}}{4 x + 7}}$

but $\frac{3 x - 2}{3 x - 2}$ is another way of writing 1 giving:

$\left(3 x + 2\right) \times 1 = \frac{\left(4 x - 7\right) \left(3 x - 2\right)}{4 x + 7}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Multiply both sides by } \textcolor{b l u e}{4 x + 7}}$

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

$\left(3 x + 2\right) \left(4 x + 7\right) = \left(4 x - 7\right) \left(3 x - 2\right) \times \frac{\left(4 x + 7\right)}{\left(4 x + 7\right)}$

But$\frac{4 x + 7}{4 x + 7}$ is another way of writing 1 giving:

$\textcolor{b r o w n}{\left(3 x + 2\right) \textcolor{b l u e}{\left(4 x + 7\right)} = \left(4 x - 7\right) \textcolor{b l a c k}{\left(3 x - 2\right)}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Multiply out the brackets}}$
color(brown)(3xcolor(blue)((4x+7))+2color(blue)((4x+7)) =4x color(black)((3x-2))-7color(black)((3x-2))

$12 {x}^{2} + 21 x + 8 x + 14 = 12 {x}^{2} - 8 x - 21 x + 14$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Collecting like terms}}$

$\left(12 {x}^{2} - 12 {x}^{2}\right) + \left(21 x + 8 x + 8 x + 21 x\right) = 14 - 14$

$0 {x}^{2} + 58 x = 0$

$\textcolor{g r e e n}{x = 0}$