Question

((2x-3)/4)-((x-2)/3)=2

Answer 1

Thank you for using parenthesis respectively.

Answer 2

\[\frac{ 2x-3 }{ 4 }-\frac{ x-2 }{ 3 }=2\]

Answer 3

Lets start by finding the LCM between \(4~,~ 3~,~ 1\) (since 2 is over 1)

Answer 4

This will help turn our function into a linear problem and make it easier to solve

Answer 5

12?

Answer 6

That's right, 12. Now we multiply all the terms by 12. \[(4)(3)\left[\frac{ 2x-3 }{ 4 }-\frac{ x-2 }{ 3 }=2\right]\]

Answer 7

when we multiply the whole equation by 12 do we multiply the numerator and the denominator too?

Answer 8

YEs

Answer 9

\[\frac{2x-3}{4} \cdot (4)(3)= 3(2x-3)\]\[-\frac{x-2}{3} \cdot (4)(3) = (4)(-x-2)\]\[2\cdot (4)(3) = 24\]

Answer 10

Do you see what I mean?

Answer 11

I don't understand how you got 3(2x-3) and 4(-x-2)

Answer 12

When I multiplied both the numerator and denominator by 12, or (4)(3), I see that the 4 in the denominator cancels with the 4 im multiplying by
\[\frac{2x-3}{\cancel{4}} \cdot \cancel{(4)}(3)= 3(2x-3)\]
Multiplying the second fraction by (4)(3), the 3 gets cancelled out the same way\[\frac{ x-2 }{ \cancel{3} } \cdot (4)\cancel{(3)} = 4(x-2)\]

Answer 13

Sorry, forget about the negative infront of the second fraction for now, we'll deal with that later.

Answer 14

Does that make more sense though?

Answer 15

yes

Answer 16

Alright

Answer 17

I got x=12.5

Answer 18

Alright let's see if that's correct

Answer 19

\[3(2x-3)-4(x-2)=24\]\[6x-9-4x+8=24\]\[2x-1=24\]\[2x=25\]\[x=\frac{25}{2} =12.5 \qquad \checkmark\] Good job.

Answer 20

thank you!