Remove parentheses and collect similar terms in the following: 21. x?— (x2y— 22) — 22+(xạy—x?). 22. x-[-{-(-x-1) —x} – 1]-1. 23. -[-2x-{-(-2 x— 1) – 2 x} – 1]–2 x. 24. x-[x+ (x −y) - {x+(y-x) — 2 y}+y]-y+. 25. If A=4 x3 – 2 x2y+3 xy2+y?, C=3 x3 – x2y+2 y?, B=4 x3 – x2y – xy2 – 3 y, D=x3 — 2 xy2 +yo, find the value of A-B+C-D. CHAPTER V SIMPLE EQUATIONS 43. Introduction. We have already seen in Chapter I how to solve such equations as 2 x–5=x+1. The steps are as follows: 2 x–5=x+1. (Given) Adding 5 to both sides gives 2 x=x+6. (Axiom I, § 9) Subtracting x from both sides (of the last equation) gives x=6. Ans. CHECK. With x=6 the left side of the given equation becomes 2X6–5, or 12–5, which is 7, while the right side becomes 6+1, which also is 7. This way of solving an equation step-by-step will always give the value of x, but in practice the work is usually much shortened in ways which we shall now explain. 44. Transposition. Let us look again at the preceding solution of the equation 2 x–5=x+1. The first step (which was to add 5 to both sides) practically amounts to getting rid of the - 5 on the left side by putting +5 on the right side. In fact, when we do this we get 2 x=x+1+5,' which reduces to 2 x=x+6. Likewise, the second step (which was to subtract « from both sides of the equation 2 x=x+6) practically amounts to getting rid of the x on the right side by placing -x on the left side. In fact, this gives -2+2 x=6, which reduces to x=6. Whenever we get rid of a term on either side of an equation in this way we are said to transpose it. Much time can be saved in making use of this idea, since we can often transpose a number of different terms all at once, thus condensing the separate steps which would be necessary if we followed the long way of Chapter I. All we have to remember is that every time a term is transposed its sign must be changed. The following examples will further illustrate this idea and its usefulness. EXAMPLE 1. Solve the equation 4 x–2=10+2 x. SOLUTION. Transposing the term –2 to the right side and the 2 x to the left side, we obtain the equation 4 x–2 x=10+2, which reduces to 2 x=12.. Hence we have x=6. Ans. (Axiom IV, $ 9) EXAMPLE 2. Solve the equation 2x-5—(6–4 x)=x—(7+3 x). SOLUTION. Removing the parentheses (see § 40) we find • 2x–5–6+4 x=2—7—3 x. Transposing all the terms containing a to the left side, and all the others to the right side, we have -x+3 x+2 x+4 x=-7+5+6. Combining terms gives - 8x=4. Dividing both sides by 8 gives x=1=1. Ans. From what we have just seen we may state the following general principle. PRINCIPLE. Any term may be transposed from one side of an equation to the other, provided its sign is changed. HISTORICAL NOTE. Our word algebra comes from the Arabic word al-gebr, meaning “to restore.” The Arabs avoided negative numbers as far as possible and the first thing they would do with such an equation as x+2=2 x-1 was to “restore" it to a form that did not contain the -1; that is, they would write the equation over in the form x+3=2 x. This is what we now call transposing the -1. ORAL EXERCISES 1. Explain the transposition in each of the following cases : (a) 3 x-1=6. (6) 4 x+2=12. (c) 2 x=13– 3 x. 3 x=6+1. 4 x=12-2. 2x+3 x=13. 2. What transposition would you suggest for each of the following in order to solve it? (a) 3 x–5=2 x+10. (d) 6=3 x–5 x+2. (6) 7 x-4 x–3 = 2 x— 2. (e) 10 x–3–4 x=5—x. (c) 9 x=5+2 x+2. (H) 2 x–5+6 x=0. 45. Cancelling Terms in an Equation. Consider the equation 2 x+7=30+7. Here 7 occurs on both sides. Hence, if we subtract it from each side (by Axiom II, § 9) it disappears entirely from the equation; that is, the equation becomes simply 2 x=30. In such a case the 7 is said to cancel from both sides. The way in which this idea is used in practice is illustrated in the following examples, where the cancelled terms have the mark / drawn through them. EXAMPLE 1. Solve the equation 3-(7—4 x)=x+(2 x+3). Solution. Removing the parentheses (see § 40) gives 3–7+4 x=x+2 x+3. Cancelling the 3 gives : 8–7+4 x=x+2 x+3, or –7+4 x=3 x. Transposing (see § 44) gives 40–3 x=7, x=7. Ans. EXAMPLE 2. Solve the equation x+(3 x+5)=5—(9— x). Solution. Removing parentheses gives x+3 x+5=5–9+x. Cancelling wherever possible gives 843 x+3=5–9+x, or 3 x=-9. Dividing by 3 gives x=-3. Ans. ORAL EXERCISES State where cancellation can take place in each of the following cases. 1. 5 x+3=20+3. 3. 5 x+x= x — 20. 2. 5x—3=20–3. 4. 4+3 x– x=6+4-x. 5. 5-(3-X)= –2 x+x-3. (Hint. First remove the parentheses.] 6. 2 x— (3 x — 3 — 7) = 3 x+(x+3). 7. (3 y–4) – [2-(y-4)] = (3 y-4) – 2. 8. 22–[2+(1-2)] = 2 z-(1+z). |