Question

Given $-\frac{7}{4}<-3t+1<-\frac{9}{5}$, find the possible value of $9t-3$.

Answer 1

$\frac{-7}{4}<-3t+1<\frac{-9}{5}$

$-1-\frac{7}{4}<-3t<\frac{-9}{5}-1$

$\frac{-4-7}{4}<-3t<\frac{-9-5}{5}$

$\frac{-11}{4}<-3t<\frac{-14}{5}$

$\frac{-11}{4\times -3}<t<\frac{-14}{5\times -3}$

$\frac{11}{12}<t<\frac{14}{15}$

$9t-3$

$t=\frac{14}{15}$

$9t-3=9\times \frac{14}{15}-3$

$=3\times 145-3$

$=\frac{42-15}{5}$

$=\frac{27}{5}$

$t=\frac{11}{12}$

$9t-3=9\times \frac{11}{12}-3$

$=\frac{3\times 11}{4}-3$

$\frac{33}{4}-3$

$\frac{33-12}{4}$

$\frac{21}{4}$

Possible values of

$9t-3=\frac{21}{4}<9t-3<\frac{27}{5}$