## EXOEXOEXO one year ago ((2x-3)/4)-((x-2)/3)=2

1. anonymous

Thank you for using parenthesis respectively.

2. EXOEXOEXO

$\frac{ 2x-3 }{ 4 }-\frac{ x-2 }{ 3 }=2$

3. anonymous

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

4. anonymous

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

5. EXOEXOEXO

12?

6. anonymous

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

7. EXOEXOEXO

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

8. anonymous

YEs

9. anonymous

$\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$

10. anonymous

Do you see what I mean?

11. EXOEXOEXO

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

12. anonymous

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$\frac{ x-2 }{ \cancel{3} } \cdot (4)\cancel{(3)} = 4(x-2)$

13. anonymous

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

14. anonymous

Does that make more sense though?

15. EXOEXOEXO

yes

16. anonymous

Alright

17. EXOEXOEXO

I got x=12.5

18. anonymous

Alright let's see if that's correct

19. anonymous

$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.

20. EXOEXOEXO

thank you!