Chapter 8.2, Problem 28IP

To determine

The solution of the equation $3b+15=27$ , and check the solution.

Answer to Problem 28IP

  $b=4$

Explanation of Solution

Given:

The equation, $3b+15=27$ .

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 given equation $3b+15=27$ for the unknown variable, 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 b on left side, first subtract 15 from both sides and then divide both sides by 3 and then simplify further as shown below,

  $\begin{array}{l}3b+15=27\\ 3b+15-15=27-15\\ 3b=12\\ \frac{3b}{3}=\frac{12}{3}\\ b=4\end{array}$

Thus, the solution of the given equation is $b=4$ .

Now, to check the solution, substitute $b=4$ in the given equation and check whether the values on both sides of equal sign is same or no. So, it gives

  $\begin{array}{l}3b+15=27\\ 3\left(4\right)+15=27\\ 12+15=27\\ 27=27\text{True Statement}\end{array}$

Since, the left hand side and right hand side are equal, so the solution is correct.