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

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

Mathematics
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Thank you for using parenthesis respectively.
\[\frac{ 2x-3 }{ 4 }-\frac{ x-2 }{ 3 }=2\]
Lets start by finding the LCM between \(4~,~ 3~,~ 1\) (since 2 is over 1)

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This will help turn our function into a linear problem and make it easier to solve
12?
That's right, 12. Now we multiply all the terms by 12. \[(4)(3)\left[\frac{ 2x-3 }{ 4 }-\frac{ x-2 }{ 3 }=2\right]\]
when we multiply the whole equation by 12 do we multiply the numerator and the denominator too?
YEs
\[\frac{2x-3}{4} \cdot (4)(3)= 3(2x-3)\]\[-\frac{x-2}{3} \cdot (4)(3) = (4)(-x-2)\]\[2\cdot (4)(3) = 24\]
Do you see what I mean?
I don't understand how you got 3(2x-3) and 4(-x-2)
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\[\frac{ x-2 }{ \cancel{3} } \cdot (4)\cancel{(3)} = 4(x-2)\]
Sorry, forget about the negative infront of the second fraction for now, we'll deal with that later.
Does that make more sense though?
yes
Alright
I got x=12.5
Alright let's see if that's correct
\[3(2x-3)-4(x-2)=24\]\[6x-9-4x+8=24\]\[2x-1=24\]\[2x=25\]\[x=\frac{25}{2} =12.5 \qquad \checkmark\] Good job.
thank you!

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