anonymous
  • anonymous
so if I have an equation with two fractions in it and two whole numbers and one of the fractions has an x next to it in order for me to get rid of those fractions I multiply by the LCD but do I also multiply the whole numbers by that?
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!
jim_thompson5910
  • jim_thompson5910
what is the equation
anonymous
  • anonymous
well like the one we just did. what if there was a whole number in there would I have to also times that by the LCD?
jim_thompson5910
  • jim_thompson5910
yes you need to multiply every term by the LCD so things balance out

Looking for something else?

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

More answers

jim_thompson5910
  • jim_thompson5910
but what is the equation?
anonymous
  • anonymous
2x+3=3/4x--2
phi
  • phi
\[ 2x+3=\frac{3}{4}x - -2 \] I would immediately write -(-2) as +2 to avoid mistakes \[ 2x+3=\frac{3}{4}x +2 \] to get rid of the 4, multiply both sides of the equation (and *all* terms) by 4. like this \[ 4\cdot 2x+ 4\cdot 3= 4\cdot \frac{3}{4}x + 4\cdot 2 \] now simplify
phi
  • phi
the first term is 4*2*x which is 8*x , usually written 8x 4*3 is 12 \[ 4 \cdot \frac{3}{4} \cdot x = \frac{4\cdot 3\cdot x}{4} \text{ or } \frac{4}{4}\cdot \frac{3x}{1}= 3x\]
phi
  • phi
the last term 4*2 is 8 you get 8x +12 = 3x +8

Looking for something else?

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