Question

Solve the linear inequality \[3x - 7 > 5x - 1\]

Answer 1

Hint: We are given an inequality in the question and we need to solve for \[x\]in the given question. While we are solving the inequalities, we will reshuffle the terms as we do in the equations but in this we have some minor differences compared to equations. We will first bring all the terms of \[x\]on one side and all the numbers on the other side and then solve both the sides individually. When we multiply the inequality with any negative number, the sign of inequality gets changed.

Complete step-by-step solution:
We are given, \[3x - 7 > 5x - 1\]
We have to solve for \[x\].
Considering \[3x - 7 > 5x - 1\]
We bring all the terms of \[x\] on the left hand side and the constant terms on the right hand side.
Subtracting \[5x\]from both the sides, we get
\[3x - 7 - 5x > 5x - 1 - 5x\]
Clubbing the variable terms together on both the sides,
\[(3x - 5x) - 7 > (5x - 5x) - 1\]
\[ \Rightarrow - 2x - 7 > 0 - 1\]
Now adding \[7\] both the sides, we get
\[ \Rightarrow - 2x - 7 + 7 > 0 - 1 + 7\]
As we know, \[ - q + q = 0\], equation becomes
\[ \Rightarrow - 2x + 0 > 0 - 1 + 7\]
Solving both sides we get,
\[ \Rightarrow - 2x > 6\]
Multiplying both the sides by \[ - 1\] and changing the inequality sign,
\[ \Rightarrow ( - 1)( - 2x) < ( - 1)6\]
\[ \Rightarrow 2x < - 6\]
Dividing both the sides by \[2\]
\[ \Rightarrow \dfrac{{2x}}{2} < \dfrac{{ - 6}}{2}\]
\[ \Rightarrow x < - 3\] which means \[x \in ( - \infty , - 3)\]
Hence, we got \[x < - 3 \Rightarrow x \in ( - \infty , - 3)\].

Note: First of all, we need to be very careful with the inequality sign. The question should be noted down correctly. Also, we forget to change the inequality sign when we multiply with any negative number or do reciprocal. When we write the range of \[x\] in interval, we need to make sure that equality sign is not there so open brackets will be used and Also, with \[\infty \] or \[( - \infty )\], always open brackets are used.