Question

What is the simplified value of: $7\dfrac{8}{9}-5\dfrac{3}{5}$ ?

Answer 1

Hint: To solve the above expression, we will first convert both the mixed numbers into their respective improper numbers. Once this is done the problem becomes easier to calculate. Now, we can subtract the second term from the first term by taking LCM of the two denominators and then subtracting the resulting values in the numerator of the expression.

Complete step by step answer:
Let us first assign some terms that we are going to use later in or solution. Let the term $7\dfrac{8}{9}$ be termed as ‘x’ and the term $5\dfrac{3}{5}$ be denoted by ‘y’, such that their difference is equal to ‘z’. Mathematically, this could be written as:
$\Rightarrow z=x-y$
Now, we will convert ‘x’ and ‘y’ into their respective improper number forms. This can be done by using the following conversion formula:
$\Rightarrow a\dfrac{b}{c}=\dfrac{c\times a+b}{c}$
Using this formula, we can write:
$\begin{align}
  & \Rightarrow 7\dfrac{8}{9}=\dfrac{9\times 7+8}{9} \\
 & \therefore 7\dfrac{8}{9}=\dfrac{71}{9} \\
\end{align}$
Therefore,
$\Rightarrow x=\dfrac{71}{9}$

For the second term, we can write:
$\begin{align}
  & \Rightarrow 5\dfrac{3}{5}=\dfrac{5\times 5+3}{5} \\
 & \therefore 5\dfrac{3}{5}=\dfrac{28}{5} \\
\end{align}$
Therefore,
$\Rightarrow y=\dfrac{28}{5}$
Now, on subtracting ‘y’ from ‘x’ to calculate the value of ‘z’, we get:
$\Rightarrow z=\dfrac{71}{9}-\dfrac{28}{5}$
Taking LCM of 9 and 5 as 45 in the denominator at the RHS of the equation, we get:
$\begin{align}
  & \Rightarrow z=\dfrac{71\times 5-28\times 9}{45} \\
 & \Rightarrow z=\dfrac{355-252}{45} \\
 & \Rightarrow z=\dfrac{103}{45} \\
 & \therefore z=2\dfrac{13}{45} \\
\end{align}$
Thus, our final result comes out to be $2\dfrac{13}{45}$.

Hence, the simplified value of, $7\dfrac{8}{9}-5\dfrac{3}{5}$ , comes out to be $2\dfrac{13}{45}$.

Note: Whenever calculating the subtraction or addition or any other operation of two mixed numbers, we should always first convert them into improper functions as it is easier to operate on improper functions. Also, if the problem is presented in mixed number format, then the final solution should be also converted into mixed form, that is, if the answer is greater than 1.