Question

Name the property used in each step:

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

Short answer

$\begin{array}{c}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{+}\mathbf{6}\mathbf{(}\mathbf{15}\mathbf{÷}\mathbf{3}\mathbf{-}\mathbf{5}\mathbf{)}\mathbf{=}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{+}\mathbf{6}\mathbf{(}\mathbf{5}\mathbf{-}\mathbf{5}\mathbf{)}\underset{\mathbf{¯}}{\mathbf{\text{Substitution}}}\mathbf{15}\mathbf{÷}\mathbf{3}\mathbf{=}\mathbf{5}\\ \mathbf{=}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{+}\mathbf{6}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\underset{\mathbf{¯}}{\mathbf{\text{AdditiveInverse}}}\mathbf{5}\mathbf{-}\mathbf{5}\mathbf{=}\mathbf{0}\\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{6}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\underset{\mathbf{¯}}{\mathbf{\text{MultiplicativeInverse}}}\mathbf{7}\mathbf{\cdot }\frac{\mathbf{1}}{\mathbf{7}}\mathbf{=}\mathbf{1}\\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{0}\underset{\mathbf{¯}}{\mathbf{\text{Multiplitionwithzero}}}\mathbf{6}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{=}\mathbf{1}\underset{\mathbf{¯}}{\mathbf{\text{Substitution}}}\mathbf{\text{1}}\mathbf{+}\mathbf{0}\mathbf{=}\mathbf{1}\end{array}$

Step-by-step solution

– Given information.

Given expression is:

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

– 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:

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

– 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)