How do you solve #2(3x-6)=3(2x-4)#?

3 Answers
Sep 15, 2016

There is no solution for #x# as it cancels out.

#color(red)("Is the question correct?")#

Explanation:

Multiply out the brackets

#6x-12=6x-12#

Add 12 to both sides

#6x=6x# #color(purple)(larr" Which is true but of no use!")#

Sep 15, 2016

#x in (-oo, oo)#

Explanation:

Given:

#2(3x-6) = 3(2x-4)#

Expand both sides to get:

#6x-12 = 6x-12#

This is true for any value of #x#, so the solution space is the whole of the Real numbers, in interval notation #(-oo, oo)#

Or did you simply want to prove the equality?

We can do that like this:

#2(3x-6) = (2*3x)-(2*6) = 6x-12 = (3*2x)-(3*4) = 3(2x-4)#

Sep 15, 2016

#x# can have any value. #x in RR#

Explanation:

#2(3x-6)= 3(2x-4)#

#6x-12 = 6x-12" "larr# both sides are the same?

#6x -6x = -12+12" "larr# follow the normal process

#0 = 0" "larr# This is TRUE, but what is #x#??

We have ended up with a true statement but there is no #x# left to solve for?

This is a special kind of equation called an identity which will be true for every value of #x#. #x# can be any Real number.

This is indicated by the TRUE statement we get.
This can be in the form #x=x#, #12=12# or similar.

The answer is therefore "x can have any value"

There are 4 different types of solutions which we can get to an equation.

#x = "a specific number"# (or numbers) as the solution(s). These are the normal equations - linear, quadratic, cubic etc.

#x = 0#
(Indicated by equations such as 23x = 12x, which at first glance seem impossible, because different quantities of x are equal to each other.) The only possible solution is if #x=0#

#x " can have any value " x in RR#
This is indicated by answers such as #0=0, " or " 8=8"# which are TRUE statements but there is no #x# term.

#x " has no value/ is undefined"#
This result is indicated by an answer such as #0=8# which is a FALSE statement but there is no #x# term.