Question

What is the sum (Two-fifths x + StartFraction 5 over 8 EndFraction) + (one-fifth x minus one-fourth)?

Answer 1

Answer:

$\frac24{x+15}{40}$

Step-by-step explanation:

We want to find the sum of $(\frac{2}{5}x+\frac{5}{8})+(\frac{1}{5}x-\frac{1}{4})$

We find the least common denominator to be 40

We collect LCM to obtain:

$\frac{2x*8+5*5+8x-10}{40}$

We simplify to get

$\frac{16x+25+8x-10}{40}$Simplify the numerator to obtain:

$\frac{24x+15}{40}$

Answer 2

Sum of $(\frac{2}{5}x+\frac{5}{8})+(\frac{1}{5}x-\frac{1}{4})$ is $\frac{3x}{5}+\frac{3}{8}$

Step-by-step explanation:

We need to find sum of (Two-fifths x + StartFraction 5 over 8 EndFraction) + (one-fifth x minus one-fourth)

Writing in mathematical form:

$(\frac{2}{5}x+\frac{5}{8})+(\frac{1}{5}x-\frac{1}{4})$

Solving

$\frac{2}{5}x+\frac{5}{8}+\frac{1}{5}x-\frac{1}{4}$

Combining like terms:

$=\frac{2}{5}x+\frac{1}{5}x+\frac{5}{8}-\frac{1}{4}\\Taking\,\,LCM\\=\frac{2x+x}{5}+\frac{5-1*2}{8}\\=\frac{3x}{5}+\frac{3}{8}$

So, sum of $(\frac{2}{5}x+\frac{5}{8})+(\frac{1}{5}x-\frac{1}{4})$ is $\frac{3x}{5}+\frac{3}{8}$