Question

Solve the following system of inequations:

$\frac{5x}{4}+\frac{3x}{8}>\frac{39}{8},\frac{2x-1}{12}-\frac{x-11}{3}<\frac{3x+1}{4}$

Answer 1

Step 1:

Solve the first linear inequation:

Consider the first inequation.

$$\begin{gathered} \Rightarrow \frac{5x}{4}+\frac{3x}{8}>\frac{39}{8} \\ \Rightarrow \frac{5x\times 2}{4\times 2}+\frac{3x}{8}>\frac{39}{8}\left[\mathrm{Taking}8\mathrm{as}\mathrm{LCM}\right] \\ \Rightarrow \frac{10x}{8}+\frac{3x}{8}>\frac{39}{8} \\ \Rightarrow \frac{10x+3x}{8}>\frac{39}{8} \\ \Rightarrow \frac{13x}{8}>\frac{39}{8} \\ \Rightarrow 13x>39\left[\mathrm{Multiply}\mathrm{by}8\mathrm{on}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow x>\frac{39}{13} \\ \therefore x>3...\left(1\right) \end{gathered}$$

Step 2:

Solve the system of inequations:

Consider linear equation $\frac{2x-1}{12}-\frac{x-11}{3}<\frac{3x+1}{4}$

Solve the inequation.

$\begin{array}{rcl}\frac{2x-1}{12}-\frac{4\left(x-11\right)}{12}& <& \frac{3\left(3x+1\right)}{12}\left[\mathrm{Taking}12\mathrm{as}\mathrm{LCM}\right]\\ & \Rightarrow & \frac{2x-1-4\left(x-11\right)}{12}<\frac{3\left(3x+1\right)}{12}\\ & \Rightarrow & \frac{2x-1-4x+44}{12}<\frac{9x+3}{12}\\ & \Rightarrow & \frac{2x-4x+44-1}{12}<\frac{9x+3}{12}\\ & \Rightarrow & \frac{-2x+43}{12}<\frac{9x+3}{12}\\ & \Rightarrow & -2x+43<9x+3\\ & \Rightarrow & -2x-9x+43<3\left[\mathrm{Subtract}9x\mathrm{on}\mathrm{both}\mathrm{sides}\right]\\ & \Rightarrow & -11x<3-43\left[\mathrm{Subtract}43\mathrm{on}\mathrm{both}\mathrm{sides}\right]\\ & \Rightarrow & -11x<-40\\ & \Rightarrow & x>\frac{-40}{-11}\left[\mathrm{Divide}\mathrm{by}-11\right]\\ \therefore x& >& \frac{40}{11}...\left(2\right)\end{array}$

Combining equation (1) and equation (2), we get the solution,

$x>\frac{40}{11}$

Thus, the required solution is $x>\frac{40}{11}$ .