Question

If $x$ belongs to a set of integers, $A$ is the solution set of $2(x-1)<3x-1$ and $B$ is the solution set of $4x-3\le 8+x$, then find $A\cap B$.

• A. $\{0,1,2\}$
• B. $\{1,2,3\}$
• C. $\{0,1,2,3\}$
• D. $\{0,2,4\}$

Answer 1

Answer: (C)

$\{0,1,2,3\}$

Consider the first inequality
$2(x-1)<3x-1$
$2x-2<3x-1$
$-1<x$ ...(i)
Consider the second inequality
$4x-3\le 8+x$
$3x\le 11$
$x\le \frac{11}{3}$
$x\le \frac{9+2}{3}$
$x\le 3+\frac{2}{3}$ but $x$ is an integer
Hence, we neglect the fractional part of $x$.
$\frac{2}{3}<1$, therefore
$x\le 3$ ...(ii)
From (i) and (ii), we have
$-1<x\le 3$
So, the solution set for the above inequalities is $\{0,1,2,3\}$.