Question

linear inequalities (level 2)
question
solve the following inequality for f. write your answer in simplest form.
10 - (6f + 3) < 3f + 1 + 6f

Explanation:

Step1: Simplify left side

First, simplify the left - hand side of the inequality \(10-(6f + 3)\). Using the distributive property \(a-(b + c)=a - b - c\), we have \(10-6f-3=7 - 6f\).

Step2: Simplify right side

Simplify the right - hand side of the inequality \(3f + 1+6f\). Combine like terms: \(3f+6f + 1=9f + 1\). So the inequality becomes \(7-6f<9f + 1\).

Step3: Add \(6f\) to both sides

To get all the \(f\) terms on one side, add \(6f\) to both sides of the inequality. \(7-6f + 6f<9f+1 + 6f\), which simplifies to \(7<15f + 1\).

Step4: Subtract 1 from both sides

Subtract 1 from both sides of the inequality: \(7-1<15f+1 - 1\), which gives \(6<15f\).

Step5: Divide both sides by 15

Divide both sides of the inequality by 15. \(\frac{6}{15}<\frac{15f}{15}\). Simplify \(\frac{6}{15}\) to \(\frac{2}{5}\) (since both numerator and denominator are divisible by 3). So we get \(\frac{2}{5}\frac{2}{5}\).

Answer:

\(f>\frac{2}{5}\)