17. 6a4-a3+2 a2+4, -4-2 a3+a? — 6 a, and 5 a4 – 4a2+2. 18. a2b2 + 2 ab3 + 4 b2, 5 a2b2 – 2 ab3 – 2 b?, and - 3 a2b2 + 2 ab3 + 362. 19. 2 H+P, 3 H-W, 3W-5 X, and 3 H+2 P+5 W. 20. 2c-7d+6n, 11 m-30-5 n, 7n-2d-8 C, 8d-3 m+10 c, 4d-3 n-8 m, m-6 n, and 2 m-3 d. 21. 4x3 – 2x2 — 7x+1, x3+3x2+53–6, 4x2 – 8x3+2–6x, 2 x3 — 2 x2 +8 x+4, and 2 x3 – 3 x2 – 2 x+1. 22. 25 +5 a+b+5 ab4 +65, a4b-2a5 +a3b2 — 2 65, a3b2 – 3 a2b3 – 4 a4b-a, and 2 aś +a4b – 2a3b2+2 a263 – 3 ab4+65. 35. Subtraction of Monomials. A monomial is an expression which contains but one term. Thus, x, 8 y, 2 m2, ab, cd8q, Im?n 34, 16 gh’ijk3 are each monomials. Compare this definition with the definitions of binomial, trinomial, etc., in § 31. To subtract one monomial from another we simply change the sign of the subtrahend and then proceed as in addition. (See $ 21.) Thus, 4 a-2 a=4 a+(-2 a)=2 a. Ans. 8 xạy-(-3 xạy)=8 xạy+(3 xạy)=11 xạy. Ans. -5 26 – 3 x5= -5 25+(-3 x5)=-8 x5. Ans. -7 a2b-(-4 a2b)=–7 a2b+4 a?b=-3 aob. Ans. In the subtraction of unlike terms we can only indicate the result. (Compare $ 30.) Thus, to subtract y from x we must write simply cậy. To subtract 3 b from 7 a we write 7 a-3 b. In all cases of the subtraction of monomials we therefore have the following rule. RULE FOR SUBTRACTING ONE MONOMIAL FROM ANOTHER. Change the sign of the subtrahend (or part subtracted) and add the result to the minuend. ORAL EXERCISES In solving an example in subtraction, the change of the sign of the subtrahend should be made mentally. The written sign should not be changed. 13. Subtract — 7 yz from – 4 yz. 14. Subtract -yk from – 7 yk. 15. Subtract ab from – ab. 20. From 18 a3 take 11 a3. 16. Subtract 21 r2 from 12 r2. 21. 3 a-(-5 a)= ? 17. From 26 xyz take – 13 xyz. 22. –6 xy-(-4 xy) = ? 18. From – 11 b take 4 b. 23. 8r-(-2 r)= ? 19. From – 9 x3y take – 7 xoy. 24. Take – 4z from – 67. State the answer in each of the following four exercises. Note that these deal with unlike terms. (See § 35.) 25. Subtract 3 b from 10 a. 28. Subtract 90 pors from abcd. For further. exercises on this topic, see Appendix, p. 295. 36. Subtraction of Polynomials. To subtract one polynomial from another, write like terms in the same column and subtract in each column separately. For example, in subtracting 3 x-8 y from 2 x+4 y the work is as follows: 2 x+4 y To check your answer, show that the sum of the remainder (or answer) and the subtrahend equals the minuend. Thus, in the example just worked, if we add – x+12 y to 3 x-8 y, we should get 2 x+4 y. Do we? If the subtrahend and the minuend contain no like terms, we can only indicate the subtraction. Thus, in subtracting 6 x+y from 2 m+n the result is 2 m+n-6 x-y. In all cases (whether the subtrahend and minuend contain like terms or not) the result can always be obtained by merely changing all signs in the subtrahend and adding the result to the minuend. Hence, we have the following rule : RULE FOR SUBTRACTING ONE POLYNOMIAL FROM ANOTHER. Change all signs in the subtrahend and add the result to the minuend. Note. Polynomials which contain different powers of some letter should first be arranged according to the descending or ascending powers of that letter. (Compare g 33.) WRITTEN EXERCISES Perform each of the following subtractions, and check your answer in each. 1. 9a+7b 2. 4 a+10b 3. 10 x+2 y 4. 4 m+4n 2 a+3 b 50+ 4b 6x+4 y 2 m+5n 5. 11 m-2 n 6. 9 a-16 b 7. 10 a, 2c 9 m- n 3 a+22 b - 16 a- 14 c 8.2 72 +8 ab3 5 p2 — 7 ab3 9. From 8 p+3 ż subtract 10 p+z. 10. From - 10 m+n subtract – 5 m-3 n. 11. From 3 ax – 5 by subtract – 4 ax+6 by. 12. From 7 abc+10 mx subtract 20 abc - 3 mx. 13. From 4 m-3 n+2p 14. From 3 a-4b+7c take 2 m-5 n- p take 6a-5b+9c 15. From a-b+c 16. From x2 – xy +y2 take 2 a+b-c take x2+xy+y? 17. From a2+2 ab+b2 18. From 8 aềb – 5 ac2+9 a c take a? — 2 ab+b2 take 3 a2b+2 ac2 — 9 aục 19. From 10 m-6 n-3 p subtract 2 n+3 p-4 m. (Hint. First rearrange the subtrahend in the form -4 m+2 n+3 p.] 20. From 6 ac+10 bd-8 s subtract 4 8-10 bd+5 ac. 21. Subtract –5 x+8 x3+7+2 x2 from -2+6 x 6 x2+4 x3. (Hint. See Note in § 36.] 22. Take 7 xy2 +9 x2y – 7 x2y2 from 4 xły–6 xy2 – 12 xy?. SOLUTION. Here the subtrahend and the minuend contain some unlike terms, and the work would therefore look as follows: 2 a+3 c+d [Hint. Since the subtrahend' here contains no terms like those in the minuend, no special arrangement of the work is necessary. Thus the answer is 4 ab+c- 5 m +2 n –3 q.] 31. From xy - 5 yz2 subtract xy2 +yz-8 x2y2 – 7 yz2. 32. From 1-2 x3+x2 subtract 24+4 x2 – 5 – 3 x. 33. From 3 x-y+z subtract the sum of x-4y+z and x+4y+z. 34. From the sum of 2 s+86-4 w and 3 s-61+2 w subtract the sum of 8 8+9t+6 w and 4 s–7t-4 w. 35. From x3 — * x2 +x-5 subtract į x3 – x2 – 2 x+1. For further exercises on this topic, see Appendix, p. 295. 37. Further Study of Equations. In $$ 6, 7, 8 we saw that numbers may be added to, or subtracted from, each side of an equation, and it was in this way that we solved the equation; that is, found the value of the unknown letter. Any equation was shown to be like a pair of balance scales, the balance being undisturbed so long as both sides of the scales were changed exactly alike, and the usual ways of doing this were stated in the axioms of $ 9. However, the only equations we learned how to solve were those having positive answers. The same axioms apply also to solving equations whose answers are negative, as we shall now show by means of two examples. |