Question

solve for f and graph the solution.

$\frac{5 - f}{7} geq 1$ or $2f + 3 < 7f - 2$

plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it.

Explanation:

Step1: Solve the first inequality

Multiply both sides of $\frac{5 - f}{7}\geq1$ by 7: $5 - f\geq7$. Then subtract 5 from both sides: $-f\geq7 - 5$, so $-f\geq2$. Multiply both sides by - 1 and reverse the inequality sign: $f\leq - 2$.

Step2: Solve the second inequality

Subtract $2f$ from both sides of $2f + 3<7f-2$: $3<7f-2f - 2$, so $3<5f - 2$. Add 2 to both sides: $3 + 2<5f$, so $5<5f$. Divide both sides by 5: $1 1$.

Step3: Graph the solution

For $f\leq - 2$, we have a closed - circle at $f=-2$ (because of $\leq$) and an arrow pointing to the left. For $f > 1$, we have an open - circle at $f = 1$ (because of $>$) and an arrow pointing to the right.

Answer:

The solution of the compound inequality is $f\leq - 2$ or $f>1$. On the number - line, we have a closed circle at $f=-2$ and an arrow to the left, and an open circle at $f = 1$ and an arrow to the right.