Chapter ISG, Problem 2.3.5P

To determine

To find correct student

Answer to Problem 2.3.5P

Both students are correct.

Explanation of Solution

Given information:

  $\frac{1}{2}n+5=\frac{17}{2}$

Calculation:

Method used by 1^st student,

  $\frac{1}{2}n+5=\frac{17}{2}$

Using subtraction property of equality,

  $\frac{1}{2}n+5-5=\frac{17}{2}-5$

Solving,

  $\frac{1}{2}n=\frac{17-10}{2}$

Using multiplication property of equality,

  $\frac{1}{2}n\times 2=\frac{17-10}{2}\times 2$

So,

  $n=17-10$

Hence, value of n is 7.

Method used by 2^nd student,

  $\frac{1}{2}n+5=\frac{17}{2}$

Using subtraction property of equality,

  $\frac{1}{2}n+5-5=\frac{17}{2}-5$

Solving,

  $\frac{1}{2}n=\frac{17-10}{2}$

Using division property of equality,

  $\frac{\frac{1}{2}n}{2}=\frac{\frac{17-10}{2}}{2}$

Solving,

  $\frac{1}{4}n=\frac{17-10}{4}$

So,

  $n=7$

Both the students are correct. It is so, if we use multiplication property, we multiply by 2, if we use division property, we divide by half. The output will be same