Answer

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Hint: We will have to solve for $x$ simply by solving the above equation, not just as an equation but with keeping inequality in mind, i.e., addition of variables and constants on both sides rather than simply transposing them.

Complete step-by-step answer:

Here, we have the given inequality of the form

$\Rightarrow 3x-7>5x-1...\text{ }\left( 1 \right)$

In the above equation (1), we could have simply transposed the RHS variables to LHS but that is inappropriate for an inequality-based equation

So, we have to add or subtract the variables on both sides to cancel out on one side, i.e.,

$\Rightarrow 3x-7>5x-1$

By adding $1$ on both sides, we get

$\begin{align}

& \Rightarrow 3x-7>5x-1 \\

& \Rightarrow 3x-7+1>5x-1+1 \\

& \Rightarrow 3x-6>5x+0 \\

& \Rightarrow 3x-6>5x \\

\end{align}$

Now, for the variable term, we can subtract $5x$ on both sides, we get

$\begin{align}

& \Rightarrow 3x-6>5x \\

& \Rightarrow 3x-6-5x>5x-5x \\

& \Rightarrow -2x-6>0 \\

\end{align}$

Now, since variable has a negative sign with it, we have to multiply $\left( -1 \right)$ on both the sides, i.e.,

$\begin{align}

& \Rightarrow -2x-6>0 \\

& \Rightarrow -\left( 2x+6 \right)>0 \\

& \Rightarrow -\left( 2x+6 \right)\left( -1 \right)<\left( 0 \right)\left( -1 \right) \\

\end{align}$

Here, in the above equation’s last step, the inequality sign got reversed with multiplication of $\left( -1 \right)$ on both sides, thus

$\begin{align}

& \Rightarrow -\left( 2x+6 \right)\left( -1 \right)<\left( 0 \right)\left( -1 \right) \\

& \Rightarrow 2x+6<0 \\

\end{align}$

Subtracting $6$ on both sides, we get

$\begin{align}

& \Rightarrow 2x+6<0 \\

& \Rightarrow 2x+6-6<0-6 \\

& \Rightarrow 2x+0<-6 \\

& \Rightarrow 2x<-6 \\

\end{align}$

Now, dividing $2$ on both sides, we get

$\begin{align}

& \Rightarrow 2x<-6 \\

& \Rightarrow \dfrac{2x}{2}<\dfrac{-6}{2} \\

& \Rightarrow x<-3,\text{ i}\text{.e}\text{.,} \\

& x\in \left( -\infty ,-3 \right) \\

\end{align}$

Hence, we can say that the value of real $x$ with given inequality is \[x\in \left( -\infty ,-3 \right)\]

Note: A simple mistake which is very common in this kind of problem is, students generally transpose the variables across the inequality like a normal equation which is not preferred, especially in case of multiplications and divisions.

Complete step-by-step answer:

Here, we have the given inequality of the form

$\Rightarrow 3x-7>5x-1...\text{ }\left( 1 \right)$

In the above equation (1), we could have simply transposed the RHS variables to LHS but that is inappropriate for an inequality-based equation

So, we have to add or subtract the variables on both sides to cancel out on one side, i.e.,

$\Rightarrow 3x-7>5x-1$

By adding $1$ on both sides, we get

$\begin{align}

& \Rightarrow 3x-7>5x-1 \\

& \Rightarrow 3x-7+1>5x-1+1 \\

& \Rightarrow 3x-6>5x+0 \\

& \Rightarrow 3x-6>5x \\

\end{align}$

Now, for the variable term, we can subtract $5x$ on both sides, we get

$\begin{align}

& \Rightarrow 3x-6>5x \\

& \Rightarrow 3x-6-5x>5x-5x \\

& \Rightarrow -2x-6>0 \\

\end{align}$

Now, since variable has a negative sign with it, we have to multiply $\left( -1 \right)$ on both the sides, i.e.,

$\begin{align}

& \Rightarrow -2x-6>0 \\

& \Rightarrow -\left( 2x+6 \right)>0 \\

& \Rightarrow -\left( 2x+6 \right)\left( -1 \right)<\left( 0 \right)\left( -1 \right) \\

\end{align}$

Here, in the above equation’s last step, the inequality sign got reversed with multiplication of $\left( -1 \right)$ on both sides, thus

$\begin{align}

& \Rightarrow -\left( 2x+6 \right)\left( -1 \right)<\left( 0 \right)\left( -1 \right) \\

& \Rightarrow 2x+6<0 \\

\end{align}$

Subtracting $6$ on both sides, we get

$\begin{align}

& \Rightarrow 2x+6<0 \\

& \Rightarrow 2x+6-6<0-6 \\

& \Rightarrow 2x+0<-6 \\

& \Rightarrow 2x<-6 \\

\end{align}$

Now, dividing $2$ on both sides, we get

$\begin{align}

& \Rightarrow 2x<-6 \\

& \Rightarrow \dfrac{2x}{2}<\dfrac{-6}{2} \\

& \Rightarrow x<-3,\text{ i}\text{.e}\text{.,} \\

& x\in \left( -\infty ,-3 \right) \\

\end{align}$

Hence, we can say that the value of real $x$ with given inequality is \[x\in \left( -\infty ,-3 \right)\]

Note: A simple mistake which is very common in this kind of problem is, students generally transpose the variables across the inequality like a normal equation which is not preferred, especially in case of multiplications and divisions.

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