Question

Solve the inequation $5\left(x-4\right)+\frac{7x}{3}<2$, where $x\in N$

Answer 1

Here, the replacement set is a set of natural numbers.

For the inequation $5\left(x-4\right)+\frac{7x}{3}<2$, $x\in N$

For the inequation $5x-20+\frac{7x}{3}<2$, $x\in N$

By adding $20$ on both sides of the above inequation, we get

$5x-20+\frac{7x}{3}+20<2+20$

$\Rightarrow 5x+\frac{7x}{3}<22$

Taking L.C.M. on the left side of the above inequation, we get

$\Rightarrow \frac{15x+7x}{3}<22$

$\Rightarrow \frac{22x}{3}<22$

By multiplying $\frac{3}{22}$ on both sides, we get

$\Rightarrow \frac{22x}{3}\times \frac{3}{22}<22\times \frac{3}{22}$

$\Rightarrow x<3$

But, only the values $1$ and $2$ from the replacement set satisfy the inequation $5\left(x-4\right)+\frac{7x}{3}<2$

Therefore, the solution set $=\left\{1,2\right\}$