Solve the inequality 1/2 (2x+1) = 1/3 x + 1/10

Question

Solve the inequality 1/2 (2x+1) >= 1/3 x + 1/10

ScanSolve AI Tutor · Accepted Answer

To solve the inequality, we will first distribute and then gather like terms.

Distribute $\frac{1}{2}$ on the left side of the inequality.
$\frac{1}{2}(2x+1) \ge \frac{1}{3}x + \frac{1}{10}$ $\frac{1}{2}(2x) + \frac{1}{2}(1) \ge \frac{1}{3}x + \frac{1}{10}$ $x + \frac{1}{2} \ge \frac{1}{3}x + \frac{1}{10}$
Subtract $\frac{1}{3}x$ from both sides and subtract $\frac{1}{2}$ from both sides to group $x$ terms and constant terms.
$x - \frac{1}{3}x \ge \frac{1}{10} - \frac{1}{2}$
Combine the $x$ terms and the constant terms.
To do this, find a common denominator for the fractions.
For the $x$ terms: $x - \frac{1}{3}x = \frac{3}{3}x - \frac{1}{3}x = \frac{2}{3}x$ .
For the constant terms: $\frac{1}{10} - \frac{1}{2} = \frac{1}{10} - \frac{5}{10} = -\frac{4}{10}$ .
$\frac{2}{3}x \ge -\frac{4}{10}$
Simplify the fraction on the right side: $-\frac{4}{10} = -\frac{2}{5}$ .
$\frac{2}{3}x \ge -\frac{2}{5}$
Multiply both sides by the reciprocal of $\frac{2}{3}$ , which is $\frac{3}{2}$ , to isolate $x$ .
$x \ge -\frac{2}{5} \times \frac{3}{2}$ $x \ge -\frac{6}{10}$ $x \ge -\frac{3}{5}$

Solve the inequality 1/2 (2x+1) >= 1/3 x + 1/10