Question

How do you solve $3x-1 < x+12$ ?

Answer 1

We are given $3x-1 < x+12$ , we are asked to solve the given problem, to do so we will learn what type of problem we have the we will use algebraic tool to find the value which will satisfy the given problem of ours. We will also use graphs to show what is the solution of $3x-1 < x+12$ , to simplify we will use addition, subtraction, division tools.

Complete step by step solution:
We are given $3x-1 < x+12$ , the $'<'$ is the symbol to denote less than so, we have $3x-1$ is less than $x+12$ as it does not contain equality, so we are given an inequality we know that solution of any problem are refer to those value which satisfies to the given problem.
In case of linear inequality it can have an infinite number of solutions, the solutions of inequality are not dependent on the order of equations.
Now as we have $3x-1 < x+12$ .
Now we will shift all variables on one side and all constant on other.
So, we first add 1 on both sides and subtract ‘x’ on both sides.
So, we get –
$3x-x-1+1 < x+12-x+1$ . So, we get –
$\Rightarrow 2x < 13$ .
Now we divide both sides by ‘2’, we get –
$\Rightarrow \dfrac{2x}{2} < \dfrac{13}{2}$ .
So, we have –
$x < \dfrac{13}{2}$
So, our solutions are all those ‘x’ which one less than $\dfrac{13}{2}=6.5$ .
We can also depict our solution on the graph.
Firstly as we have just 1 variable so our graph will be x-axis.
Now we first find $x=\dfrac{13}{2}=6.5$ on the x-axis .

Now as we get that solution are $x < 6.5$ have less than in strict (mean that 6.5 is not included)
So, here solution do not include 6.5,
Solutions are all value less than 6.5.
Values on the left of 6.5 denote the value less than 6.5.
Hence our solution are =

Note:
When we solve inequality, we need to be known that when we divide inequality by positive number there is no affection inequality but if we divide the inequality by negative number then the sign of inequality get reversed less than ‘<’ change to ‘>’ greater than or greater than $\left( > \right)$ change to less than $\left( < \right)$ .
We can subtract term or add term; they do not affect our involving inequalities.