Question

Find the maximum value of $x,$ if $\frac{2x}{3}+7\le 17$

Answer 1

Given, $\frac{2x}{3}+7\le 17$

Step 1: Subtract 7 from both sides of the inequality in order to remove the 7 on the left hand side.
$\Rightarrow \frac{2x}{3}+7-7\le 17-7\Rightarrow \frac{2x}{3}\le 10$

Step 2: Now, multiply both sides of the inequality by 3 in order to remove the 3 on the left hand side.
$\Rightarrow \frac{2x}{3}\times 3\le 10\times 3\Rightarrow 2x\le 30$

Step 3: Now, divide both sides of the inequality by 2 in order to remove the 2 on the left hand side.
$\Rightarrow \frac{2x}{2}\le \frac{30}{2}\Rightarrow \frac{2\times x}{2}\le \frac{2\times 15}{2}$
$\Rightarrow x\le 15$

Note: Since we are multiplying and dividing by positive number, the direction of the inequality does not change

So the value(s) of $x$ which are less than or equal to $15$ satisfy the given inequality.

Hence, the maximum value of $x$ is $15.$