A linear equation can be written as $ax+b=0$, where $a$ and $b$ are constants. Write the equation $3(x-1)=5(2x+1)$ in the form of $ax+b=0$.

Sasha solves a linear equation.

Her work is shown below.

$2x-3=\frac{x}{2}-5$

$2\left(2x-3\right)=x-5$...Step $1$

$4x-6=x-5$...Step $2$

$3x=1$...Step $4$

$x=\frac{1}{3}$

Is Sasha's solution correct? If not, in which step did Sasha make an error.

Which of the following statements is CORRECT?

Statement-$1$: $x=\frac{1}{2}$ is the solution of $\frac{2x-3}{4}-\frac{2x-1}{2}=\frac{x-2}{3}$

Statement -$2$: $x=\frac{63}{2}$ is the solution of $\frac{2x-17}{2}-(x-\frac{x-1}{3})=12$

In the following question two equations I and II are given You have to solve both the equations and give answer:

$$\begin{gathered} \mathrm{I}. \frac{3}{\sqrt{\mathrm{x}}}+\frac{4}{\sqrt{\mathrm{x}}}=\sqrt{\mathrm{x}} \\ \mathrm{II}. {\mathrm{y}}^3-\frac{{\left(7\right)}^{{\displaystyle \frac{7}{2}}}}{\sqrt{\mathrm{y}}}=0 \end{gathered}$$