Question

Solve $\frac{4x}{3}-\frac{9}{4}<x+\frac{3}{4},\frac{7x-1}{3}-\frac{7x+2}{6}>x$

• A. $\left(-\infty ,4\right)\cup \left(9,\infty \right)$
• B. $\left(4,9\right)$
• C. $\left(-\infty ,\infty \right)$
• D. $\left(-\infty ,5\right)$

Answer 1

Answer: (B)

$\left(4,9\right)$

Given inequality

$\frac{4x}{3}-\frac{9}{4}<x+\frac{3}{4}\dots \left(1\right)$ and $\frac{7x-1}{3}-\frac{7x+2}{6}>x\dots \left(2\right)$

Solving the above inequality (1) we get,

$\frac{4x}{3}-\frac{9}{4}-\frac{3}{4}<x$

$\frac{16x-27-9}{12}<x$

$16x-36<12x$

$16x-12x<36$

$4x<36$

$x<\frac{36}{4}$

or $x<9....\left(3\right)$

Now, solving inequality (2)

$\frac{14x-2-7x-2}{6}>x$

$\frac{7x-4}{6}>x$

$7x-4>6x$

$7x-6x>4$

$x>4....\left(4\right)$

Hence, from equation (3) and (4) thedomain of x is given by

$(4,9)$