
Solve the given inequality for real $x$, $3x-7>5x-1$.
Answer
606.9k+ views
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.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

