



 
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
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 



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