Question

Find two mixed numbers so that the sum is 7 2/8 and the difference is 2 4/8

Answer 1

so let's name the numbers x and y.

we know that their sum, x+y is $7\frac{2}{8}$

and that that their difference, x-y is $2\frac{4}{8}$

so, from the last one we can find out x:

x-y=$2\frac{4}{8}$
x=$2\frac{4}{8}$+y


we can substitute this into the first one:

$2\frac{4}{8}$+y+y = $7\frac{2}{8}$

$2\frac{4}{8}$+2y = $7\frac{2}{8}$

2y=$4\frac{6}{8}$ (7-2=5, but you have to "borrow" one because the fraction of the number from which you substract is smaller than of the other number)

so y=$2\frac{3}{8}$

then x will be
x=$2\frac{4}{8}$+$2\frac{3}{8}$
x=$4\frac{7}{8}$