Question

Show that $2(x-1)+3(x-2)\le 5(x+1)$, $x\in \mathbb{R}$.

Answer 1

Both the sides of the inequality are linear equations in one variable. To solve this question, we will first rearrange the inequality by shifting terms from one side to another. Then we will simplify the equation using basic arithmetic operations. After simplifying the equation, we will be able to prove the inequality.

Complete step-by-step solution:
The given equation that we have to prove is $2(x-1)+3(x-2)\le 5(x+1)$. We will first rearrange the terms. We will shift the terms on the LHS to RHS. So now we have to prove the following equation,
$0\le 5(x+1)-2(x-1)-3(x-2)$.
Let $f(x)=5(x+1)-2(x-1)-3(x-2)$. Now, we will simplify this function. We have the following equation,
$\begin{aligned} & f(x)=5(x+1)-2(x-1)-3(x-2) \\\ & =5x+5-2x+2-3x+6 \end{aligned}$
We will now collect the variable terms and constant terms together. So we get the following equation,

& f(x)=(5x-2x-3x)+(5+2+6) \\\ & =0x+13 \end{aligned}$$ **After simplifying the function, we have $f(x)=13$. It is clear that $0\le f(x)$. Hence, we have proved the inequality.** **Note:** This question can be solved in other ways. One way is to plot the graphs for the functions on the LHS and the RHS and then compare the graphs for the two functions. Both functions are linear, so their graphs will be two straight lines, and hence they can be compared easily by looking at their graphs. There is one more way to solve this question, we can check the values of both, the LHS and the RHS, for the same input $x\in \mathbb{R}$, and compare the values of the output.