Chapter 8.8, Problem 19IP

To determine

Find value of the unknown variable from the given equation.

Answer to Problem 19IP

  $y=2$

Explanation of Solution

Given:

The Equation: $12y+5(y-6)=4$

Concept Used:

In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value.

Here, for example, 5x + 9 is the expression on the left-hand side, which is equal to the expression 24 on the right-hand side. i.e. 5x+9 = 24 is an equation.

Calculation:

Addition or Subtraction Property of Equality:

If $\text{a}\text{=}\text{b}$ , then $\text{a}\text{+}\text{c}\text{=}\text{b}\text{+}\text{c}$ or If $\text{a}\text{=}\text{b}$ , then $\text{a}-\text{c}\text{=}\text{b}-\text{c}$

The property that states that if you add or subtract the same number to both sides of an equation, the sides remain equal (i.e., the equation continues to be true.)

Multiplication and Division Properties of Equality:

If $\text{a}\text{=}\text{b}$ , then $\text{a}·\text{c}=\text{b}·\text{c}$

If $\text{a}\text{=}\text{b}$ , then $\text{a}÷\text{c}=\text{b}÷\text{c}$ , where c is not equal to 0.

In other words, if two expressions are equal to each other and you multiply or divide (except for 0) the exact same constant to both sides, the two sides will remain equal. 

Given the Equation: $12y+5(y-6)=4$

Step 1: Use the distributive method of multiplication to open the parenthesis.

Rule: $a·(b\pm c)=a·b\pm a·c$

  $\begin{array}{l}12y+5(y-6)=4\\ 12y+5y-30=4[a·(b\pm c)=a·b\pm a·c]\\ 17y-30=4\\ 17y-30+30=4+30\text{[}\text{Add}30\text{to}\text{both}\text{sides}\text{]}\\ 17y=34\\ \frac{17y}{17}=\frac{34}{17}\text{[}\text{Divide}\text{both}\text{sides}\text{by}17\text{]}\\ y=2\end{array}$  $\begin{array}{l}\text{Check}\text{the}\text{solution}\text{when}y=2\\ 12y+5(y-6)=4\\ 12·2+5(2-6)=4\\ 24-20=4\\ 4=4[\text{Checked}]\end{array}$

Thus, the solution of the equation $12y+5(y-6)=4$is $y=2$