Hi Ali,

For the example you are using the sum is 11 and the difference is 3, Stephen's technique says add the two numbers (11 + 3 = 14) and this is twice one of the two numbers. Thus one of the two numbers is $\large \frac{14}{2} \normalsize = 7.$ But the sum of the two numbers is 11 so the second number must be $11 - 7 = 4.$

Let's look at another example which might seem a little strange at first. Suppose the sum of the two numbers is 9 and the difference is 15. Again, Stephens technique says that the total of the sum and difference ( 9 + 15 = 24) is twice one of the numbers. Hence one of the two numbers being sought is $\large \frac{24}{2} \normalsize = 12.$ But the sum of the two numbers sought is 9 so the other number is $9 - 12 = -3.$ Make sure you check my answer.

Penny

Mike wrote:

I was reviewing your answer.
Suppose the sum of the two numbers is 9 and the difference is 15.
Again, Stephens technique says that the total of the sum and
difference ( 9 + 15 = 24) is twice one of the numbers.
Hence one of the two numbers being sought is 24/2=12.
But the sum of the two numbers sought is 9.
So the other number is 9 − 12 = −3
Question: How do you solve for -3 when all we have are 24, 12 without knowing 9.

Suppose the two unknown numbers are $a$ and $b$ and you know that

\[a + b = 9 \mbox{ and } a - b = 15.\]

If you subtract the difference $a - b$ from the sum $a + b$ you get

\[a + b - (a - b) = a + b - a + b = 2b.\]

Thus in the example where $a + b = 9$ and $a - b = 15,$

\[a + b - (a - b) = 9 - 15 = -6.\]

Hence $2b = -6$ and $b = -3.$

Penny