By Oujilide Jiheyuanben

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**Extra resources for The Thirteen Books of the Elements**

**Sample text**

1. 19 − 12 − 8 − 2 solution Here are the possibilities: Best way to learn: Carefully read the section of the textbook, then do all the odd-numbered exercises (even if they have not been assigned) and check your answers here. If you get stuck on an exercise, reread the section of the textbook—then try the exercise again. If you are still stuck, then look at the workedout solution here. 19(−12 − 8 − 2) = −418 19 − (12 − 8) − 2 = 13 19(−12 − 8) − 2 = −382 19 − (12 − 8 − 2) = 17 19(−12) − 8 − 2 = −238 19 − 12 − 8(−2) = 23 (19 − 12) − 8 − 2 = −3 19 − 12(−8) − 2 = 113 19 − 12 − (8 − 2) = 1 19 − 12(−8 − 2) = 139 Other possible ways to insert one pair of parentheses lead to values already included in the list above.

Putting all this together, we have used the distributive property (twice) to transform 2(3m + x) + 5x into the simpler expression 6m + 7x. 2 Algebra of the Real Numbers 11 One of the most common algebraic manipulations involves expanding a product of sums, as in the following example. Expand (a + b)(c + d). example 3 solution Think of (c + d) as a single number and then apply the distributive property to the expression above, getting (a + b)(c + d) = a(c + d) + b(c + d). Now apply the distributive property twice more, getting (a + b)(c + d) = ac + ad + bc + bd.

3. 6 + 3 · 4 + 5 · 2 solution (x + y + z)2 = (x + y + z)(x + y + z) solution Here are the possibilities: (6 + 3 · 4 + 5 · 2) = 28 6 + (3 · 4 + 5) · 2 = 40 (6 + 3) · 4 + 5 · 2 = 46 = x(x + y + z) + y(x + y + z) + z(x + y + z) = x 2 + xy + xz + yx + y 2 + yz + zx + zy + z2 = x 2 + y 2 + z2 + 2xy + 2xz + 2yz 6 + 3 · (4 + 5 · 2) = 48 6 + 3 · (4 + 5) · 2 = 60 Other possible ways to insert one pair of parentheses lead to values already included in the list above. For example, 13. (x + 1)(x − 2)(x + 3) solution (x + 1)(x − 2)(x + 3) = (x + 1)(x − 2) (x + 3) (6 + 3 · 4 + 5) · 2 = 46.

### The Thirteen Books of the Elements by Oujilide Jiheyuanben

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