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

2 Answers
Jan 10, 2016

Answer:

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#

Jan 13, 2016

Answer:

.
#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)#