manuel ate 1/3 of the crakers on the plate then his brother a 1/4 of craker on the plate. there where 5 left on the plate how many where left.

Answer 1

Hi, Zoe. I think you left a verb out of the problem: I believe Manuel's brother left 1/4 of the crackers on the plate. Let me know if I got this wrong.

The trick is recognizing when to use what is left and when to use what was eaten.

If Manuel ate 1/3 of the crackers, then 2/3 of the crackers were left for Manuel's brother. Make c=the number of crackers at the beginning of the problem.

2x is the number of crackers Manuel's brother has to feast on

3

Since I am assuming that Manuel's brother left 1/4 of the remaining crackers behind, this would be

1 times 2x = 5

4 3

and all of that equals the 5 leftover crackers.

To handles fractions like this in algebra, multiply by the reciprocal (which is the fraction upside down.)

1 times 2 = 2

4 3 12

So we need to multiply both sides of the equation by 12/2, because that is the fraction 2/12 upside down.

5 times 12/2 = 60/2 = 30.

So there should be 30 crackers on the plate at the beginning of the problem, but only if I guessed correctly about what was missing from the problem.