EXOEXOEXO
  • EXOEXOEXO
((2x-3)/4)-((x-2)/3)=2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jhannybean
  • Jhannybean
Thank you for using parenthesis respectively.
EXOEXOEXO
  • EXOEXOEXO
\[\frac{ 2x-3 }{ 4 }-\frac{ x-2 }{ 3 }=2\]
Jhannybean
  • Jhannybean
Lets start by finding the LCM between \(4~,~ 3~,~ 1\) (since 2 is over 1)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.