Question

1. $5k>40$
2. $\frac{t}{3}>8$
3. $10leq - 3k - 8$
4. $7x + 9geq - 26$
5. $3(5a + 2)geq51$
6. $-4(1 + 3p)< - 64$
7. $-3 + 2p<4p + 3$
8. $-9 + 4ygeq1 + 2y$
9. you have at most $3.65 to make copies. each copy costs $0.25. write and solve an inequality that represents the number of copies you can make.
10. the maximum speed limit for a school bus is 55 mph. write and solve an inequality that represents the number of hours it takes to travel 165 miles in a school bus.

Explanation:

Step1: Solve 5k > 40

Divide both sides by 5: $k>\frac{40}{5}=8$.

Step2: Solve $\frac{t}{3}>8$

Multiply both sides by 3: $t > 8\times3=24$.

Step3: Solve $10\leq - 3k - 8$

Add 8 to both sides: $10 + 8\leq-3k$, so $18\leq - 3k$. Then divide both sides by - 3 and reverse the inequality sign: $k\leq - 6$.

Step4: Solve $7x + 9\geq - 26$

Subtract 9 from both sides: $7x\geq-26 - 9=-35$. Then divide both sides by 7: $x\geq - 5$.

Step5: Solve $3(5a + 2)\geq51$

First, distribute the 3: $15a+6\geq51$. Then subtract 6 from both sides: $15a\geq51 - 6 = 45$. Finally, divide both sides by 15: $a\geq3$.

Step6: Solve $-4(1 + 3p)<-64$

First, distribute the - 4: $-4-12p<-64$. Then add 4 to both sides: $-12p<-64 + 4=-60$. Divide both sides by - 12 and reverse the inequality sign: $p > 5$.

Step7: Solve $-3 + 2p<4p+3$

Subtract 2p from both sides: $-3<4p - 2p+3$, so $-3<2p + 3$. Then subtract 3 from both sides: $-3-3<2p$, so $-6<2p$. Divide both sides by 2: $p>-3$.

Step8: Solve $-9 + 4y\geq1+2y$

Subtract 2y from both sides: $-9 + 4y-2y\geq1$, so $-9 + 2y\geq1$. Then add 9 to both sides: $2y\geq1 + 9=10$. Divide both sides by 2: $y\geq5$.

Step9: Solve the copy - making problem

Let the number of copies be $n$. The cost of $n$ copies is $0.25n$. Since you have at most $3.65$, the inequality is $0.25n\leq3.65$. Divide both sides by 0.25: $n\leq\frac{3.65}{0.25}=14.6$. Since you can't make a fraction of a copy, $n\leq14$.

Step10: Solve the school - bus problem

Let the number of hours be $h$. The speed is 55 mph and the distance is 165 miles. Using the formula $d = vt$ (distance = speed×time), we have $55h\geq165$. Divide both sides by 55: $h\geq3$.

Answer:

1. $k > 8$
2. $t>24$
3. $k\leq - 6$
4. $x\geq - 5$
5. $a\geq3$
6. $p > 5$
7. $p>-3$
8. $y\geq5$
9. $n\leq14$
10. $h\geq3$