Question

1. ten more than the quotient of a number and 3 equals -21. 6. the sum of 9 and the quotient of a number and 7 is 11.

Explanation:

Step1: Translate problem 5 into an equation

Let the number be $x$. The quotient of the number and 3 is $\frac{x}{3}$, and ten more than that is $\frac{x}{3}+ 10$. So the equation is $\frac{x}{3}+10=-21$.

Step2: Solve the equation for problem 5

Subtract 10 from both sides: $\frac{x}{3}=-21 - 10=-31$.
Multiply both sides by 3: $x=-31\times3=-93$.

Step3: Translate problem 6 into an equation

Let the number be $y$. The quotient of the number and 7 is $\frac{y}{7}$, and the sum of 9 and that is $9+\frac{y}{7}$. So the equation is $9+\frac{y}{7}=11$.

Step4: Solve the equation for problem 6

Subtract 9 from both sides: $\frac{y}{7}=11 - 9 = 2$.
Multiply both sides by 7: $y=2\times7 = 14$.

Answer:

1. $x=-93$
2. $y = 14$