Chapter 8.4, Problem 47HP

To determine

Write and solve an eqation to find Joaquin’s age.

Answer to Problem 47HP

  $21\text{years}$

Explanation of Solution

Given:

Joaquin is 4 years older thanBercky. Three times the sum of their ages is114.

Concept Used:

• To get rid of a number in addition from one side, subtract the same number from both sides of equal sign.
• To get rid of a number in subtraction from one side, add the same number both sides of equal sign.
• To get rid of a number in multiplication from one side, divide the same number from both sides of equal sign.
• To get rid of a number in division from one side, multiply the same number both sides of equal sign.

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to solve the equation to find Joaquin’s age, let j be the age of Joaquin, three times the sum of their ages is 114 as given in the question and Joaquin is 4 years older than Bercky, so j-4 is Bercky’s age , so the equation is as shown below:

  $\begin{array}{l}3(j+j-4)=114\\ 3(2j-4)=114\end{array}$

Now, to solving the equation isolate the variable term b on one side by performing some basic algebraic operations to get rid of the other numbers and terms associated with it.

Here to isolate j , first divide both sides by 3,then add 4 both sides and then again divide both sides by 2 and simplify further as shown below,

  $\begin{array}{l}3(2j-4)=114\\ \frac{3(2j-4)}{3}=\frac{114}{3}\\ 2j-4=38\\ 2j-4+4=38+4\\ 2j=42\\ \frac{2j}{2}=\frac{42}{2}\\ j=21\end{array}$

So, Joaquin’s age is 21 years.