Question

Solve the system of inequality $5x-7<3(x+3),1-\frac{3x}{2}\ge x-4$

• A. $x\in (-\infty ,2]$
• B. $x\in (8,\infty )$
• C. No solution
• D. $x\in (-\infty ,8)$

Answer 1

The correct option is A $x\in (-\infty ,2]$

We have,
$5x-7<3(x+3)$
$\Rightarrow 5x-7<3x+9$
$\Rightarrow 5x-3x<9+7$
$\Rightarrow 2x<16\Rightarrow x<8$
$\Rightarrow x\in (-\infty ,8)\cdots \left(A\right)$

And, $1-\frac{3x}{2}\ge x-4$
$\Rightarrow \frac{2-3x}{2}\ge x-4\Rightarrow 2-3x\ge 2x-8$
$\Rightarrow 2+8\ge 2x+3x$
$\Rightarrow 10\ge 5x\Rightarrow 5x\le 10\Rightarrow x\le \frac{10}{5}$
$\Rightarrow x\le 2\Rightarrow x\in (-\infty ,2]\cdots \left(B\right)$
Now, the solution set will be the intersection of both $\left(A\right)\&\left(B\right)$ as:


$\Rightarrow x\in (-\infty ,2]$

Hence the correct answer is option (b).