Elements of Algebra

208 ELEMENTS SECT. IV. question, by requiring two numbers, whose sum may be a, and the difference b. Let the greater of the two numbers be expressed by $x$, and the less by $y$; we shall then have $x + y = a$, and $x - y = b$. Here the first equation gives $x = a - y$, and the second $x = b + y$. Therefore, $a - y = b + y$; $a = b + 2y$; $2y = a - b$; lastly, $y = \frac{a-b}{2}$, and, consequently, $x = a - y = a - \frac{a-b}{2} = \frac{a+b}{2}$. Thus, we find the greater number, or $x$, is $\frac{a+b}{2}$, and the less, or $y$, is $\frac{a-b}{2}$; or, which comes to the same, $x = \frac{1}{2}a + \frac{1}{2}b$, and $y = \frac{1}{2}a - \frac{1}{2}b$. Hence we derive the following theorem: When the sum of any two numbers is a, and their difference is b, the greater of the two numbers will be equal to half the sum *plus* half the difference; and the less of the two numbers will be equal to half the sum *minus* half the difference. 611. We may resolve the same question in the following manner: Since the two equations are, $x + y = a$, and $x - y = b$; if we add the one to the other, we have $2x = a + b$. Therefore $x = \frac{a+b}{2}$. Lastly, subtracting the same equations from each other, we have $2y = a - b$; and therefore $y = \frac{a-b}{2}$. 612. Question 3. A mule and an ass were carrying burdens amounting to several hundred weight. The ass complained of his, and said to the mule, I need only one hundred weight of your load, to make mine twice as heavy as yours; to which the mule answered, But if you give me a hundred weight of yours, I shall be loaded three times as much as you will be. How many hundred weight did each carry?