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

Question

Eva Abramson · Accepted Answer

x = 0

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#

Kennedy Adams · Answer

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#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.

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)#

Nathan Axel · Answer

undefined To solve the equation (3x+2)/(3x-2)=(4x-7)/(4x+7), we can cross-multiply to eliminate the fractions. This gives us (3x+2)(4x+7) = (4x-7)(3x-2). Expanding both sides of the equation, we get 12x^2 + 21x + 8x + 14 = 12x^2 - 8x - 21x + 14. Simplifying further, we have 12x^2 + 29x + 14 = 12x^2 - 29x + 14. By subtracting 12x^2 from both sides, we get 29x + 14 = -29x + 14. Next, we can add 29x to both sides, resulting in 58x + 14 = 14. Subtracting 14 from both sides, we have 58x = 0. Finally, dividing both sides by 58, we find that x = 0.