How do you solve $\frac{3 x + 2}{3 x - 2} = \frac{4 x - 7}{4 x + 7}$ $(3x+2)/(3x-2)=(4x-7)/(4x+7)$ ?

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How do you solve $\frac{3 x + 2}{3 x - 2} = \frac{4 x - 7}{4 x + 7}$ $(3x+2)/(3x-2)=(4x-7)/(4x+7)$ ?

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Answer 1 · 2016-01-10T18:15:40

x = 0

Explanation:

Awesome fact :

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

then

$a d = b c$ $ad = bc$

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

you have

$\frac{3 x + 2}{3 x - 2} = \frac{4 x - 7}{4 x + 7}$ $(3x+2)/(3x-2)=(4x-7)/(4x+7)$ $\Rightarrow$ $=>$

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

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

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

then :

$7 v = 2 u$ $7v = 2u$

$21 x = - 8 x$ $21x = -8x$

$29 x = 0$ $29x = 0$

$x = 0$ $x = 0$

Answer 2 · 2016-01-13T10:30:19

.
$x = 0$ $color(green)(x=0)$ ..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: $\frac{3 x + 2}{3 x - 2} = \frac{4 x - 7}{4 x + 7}$ $color(brown)( (3x+2)/(3x-2) = (4x-7)/(4x+7)$

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

$(3 x + 2) \times \frac{(3 x - 2)}{(3 x - 2)} = \frac{(4 x - 7) (3 x - 2)}{4 x + 7}$ $color(brown)( (3x+2)xx color(blue)((3x-2))/((3x-2))=((4x-7)color(blue)((3x-2)))/(4x+7))$

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

$(3 x + 2) \times 1 = \frac{(4 x - 7) (3 x - 2)}{4 x + 7}$ $(3x+2) xx 1= ((4x-7)(3x-2))/(4x+7)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Multiply both sides by 4 x + 7$ $color(green)("Multiply both sides by "color(blue)(4x+7))$

$(3 x + 2) \times (4 x + 7) = \frac{(4 x - 7) (3 x - 2)}{4 x + 7} \times (4 x + 7)$ $color(brown)((3x+2) color(blue)(xx(4x+7))= ((4x-7)(3x-2))/(4x+7)color(blue)(xx(4x+7))$

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

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

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

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

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

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

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

$x = 0$ $color(green)(x=0)$

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