Chapter 6.3, Problem 42PFA

To determine

To find the numbers

Answer to Problem 42PFA

$x=68$ and $y=52$

Explanation of Solution

Given:

Sum of two numbers: 120

Three times their difference: 48

Calculation:

Let x and y be the two numbers.

Given that the sum of the numbers is 120,

  $x+y=120\to (1)$

Also, three times their difference is 48. Then,

  $3x-3y=48\to (2)$

Multiplying (1) with 3,

  $3x+3y=360\to (3)$

Adding equation (1) and equation (2),

  $\begin{array}{l}\Rightarrow (3x+3y)+(3x-3y)=360+48\\ \Rightarrow 3x+3y+3x-3y=408\\ \Rightarrow 6x=408\\ \Rightarrow x=68\end{array}$

Now, substitute value of x in equation (1).

  $\begin{array}{l}\Rightarrow 68+y=120\\ \Rightarrow y=120-68\\ \Rightarrow y=52\end{array}$

Conclusion:

Therefore, the required numbers are $x=68$ and $y=52$ .