Question

If 23/8, 11/2, 9/6, and 15/7 are placed in order from least to greatest, which would be first?

Answer 1

Answer:

$\frac{9}{6}$

Step-by-step explanation:

The given fractions are;

$\frac{23}{8},\frac{11}{2},\frac{9}{6},\frac{15}{7}$

The least common denominator is $168$.

We express all the fractions in equivalent form with the LCD as the denominator.

$\frac{23}{8}=\frac{21\times23}{21\times8}=\frac{483}{168}$

$\frac{11}{2}=\frac{84\times11}{84\times2}=\frac{924}{168}$

$\frac{9}{6}=\frac{28\times9}{28\times6}=\frac{252}{168}$

$\frac{15}{7}=\frac{24\times15}{24\times7}=\frac{360}{168}$

We now compare the equivalent forms and arrange from the least to the greatest.

$\frac{252}{168}\:<\:\frac{360}{168}\:<\:\frac{483}{168}\:<\:\frac{924}{168}$

This implies that,

$\frac{9}{6}\:<\:\frac{15}{7}\:<\:\frac{23}{8}\:<\:\frac{11}{2}$

Therefore the first would be $\frac{9}{6}$.