Answer
Verified
479.1k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE