Q. 1 BCYP - Algebra 1

Question

Name the property used in each step:

$\begin{array}{c}7\cdot \frac{1}{7}+6\left(15÷3-5\right)=7\cdot \frac{1}{7}+6\left(5-5\right)\text{ \_\_\_\_\_\_\_}\\ =7\cdot \frac{1}{7}+6\left(0\right)\text{ \_\_\_\_\_\_\_}\\ =1+6\left(0\right)\text{ \_\_\_\_\_\_\_}\\ =1+0\text{ \_\_\_\_\_\_\_}\\ =1\text{ \_\_\_\_\_\_\_}\end{array}$

Step-by-Step Solution

Answer

$\begin{array}{c}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{+}\mathbf{6}\mathbf{(}\mathbf{15}\mathbf{÷}\mathbf{3}\mathbf{-}\mathbf{5}\mathbf{)}\mathbf{=}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{+}\mathbf{6}\mathbf{(}\mathbf{5}\mathbf{-}\mathbf{5}\mathbf{)}\underset{\mathbf{¯}}{\mathbf{\text{Substitution}}}\mathbf{15}\mathbf{÷}\mathbf{3}\mathbf{=}\mathbf{5}\\ \mathbf{=}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{+}\mathbf{6}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\underset{\mathbf{¯}}{\mathbf{\text{Additive Inverse}}}\mathbf{5}\mathbf{-}\mathbf{5}\mathbf{=}\mathbf{0}\\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{6}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\underset{\mathbf{¯}}{\mathbf{\text{Multiplicative Inverse}}}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{=}\mathbf{1}\\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{0}\underset{\mathbf{¯}}{\mathbf{\text{Multiplition with zero}}}\mathbf{6}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{=}\mathbf{1}\underset{\mathbf{¯}}{\mathbf{\text{Substitution}}}\mathbf{\text{ 1}}\mathbf{+}\mathbf{0}\mathbf{=}\mathbf{1}\end{array}$

Step 1 – Given information.

Given expression is:

$\begin{array}{c}7\cdot \frac{1}{7}+6\left(15÷3-5\right)=7\cdot \frac{1}{7}+6\left(5-5\right)\text{ \_\_\_\_\_\_\_}\\ =7\cdot \frac{1}{7}+6\left(0\right)\text{ \_\_\_\_\_\_\_}\\ =1+6\left(0\right)\text{ \_\_\_\_\_\_\_}\\ =1+0\text{ \_\_\_\_\_\_\_}\\ =1\text{ \_\_\_\_\_\_\_}\end{array}$

Step 2 – Write the concept.

The substitution property for multiplication, addition and subtraction is: 

$\begin{array}{c}a\cdot b=c\\ a\pm b=c\end{array}$

Property of multiplicative inverse is: 

$a\cdot \frac{1}{a}=1$

Property of additive inverse is:

$a+\left(-a\right)=0$

Step 3 – Write the property.

Since,

$\begin{array}{c}7\cdot \frac{1}{7}+6\left(15÷3-5\right)=7\cdot \frac{1}{7}+6\left(5-5\right)\text{ \_\_\_\_\_\_\_}\\ =7\cdot \frac{1}{7}+6\left(0\right)\text{ \_\_\_\_\_\_\_}\\ =1+6\left(0\right)\text{ \_\_\_\_\_\_\_}\\ =1+0\text{ \_\_\_\_\_\_\_}\\ =1\text{ \_\_\_\_\_\_\_}\end{array}$

In 1^st step,

$15÷3=5$ (Substitution Property) 

In 2^nd step,

$5+\left(-5\right)=0$ (Property of Additive inverse) 

In 3^rd step,

$7\cdot \frac{1}{7}=1$ (Property of multiplicative inverse) 

In 4^th step,

$6\left(0\right)=0$ (Multiplication with 0) 

In 5^th step,

$1+0=1$ (Substitution Property)