long as the denominator remains, there is no fear of mistake; but when the denominator is taken away by multiplication, unless the numerator be enclosed in brackets, or the signs at once changed, errors will be sure to occur. Thus, a d 46. If multiplied by b, becomes a d b -- a + d = (a + d) = 46b, or - d = 46b. If we represent the operation in If each side of this equation be multiplied by 8, we have 384 (12 + 4) = 368 = 384 · 12 — 4 = 368. 506. EXAMPLES. By multiplying by 12, we have (6 + 4 + 3) a = 120 13x120 Therefore, 6 is the value of x. 4. Given (12 + x)2 = 2 + 2 to find x. By multiplying by 10. 40x 5 5x+5=10x + 4x4 +240 ,, transposing the terms, (40 — 5 — 10 — 4) ≈ — 240 — 4 — 5 ,, collecting, ... 21x231 21 10. There is a number to which, if we add and of itself, the sum will be 132, what is that number? 11. A person sold goods for £26, and cleared of what they cost; what did he buy them for ? 12. What sum placed at interest at 5 per cent. will amount to £405 in 5 years? Note. As every £100 gains £5, in the year, and 100- 20. Therefore the sum placed at interest, whatever it be, will gain of itself every year; and in 5 years will gain of itself, which, added to itself will give £405. 12. A man and his son being engaged to dig a piece of ground, the father could dig it in 8 days, the son 12 days; in what time will they dig it if they work together? 13. A person bought a score of oxen for £384, and sold them for of what they cost him, what did he sell them for ? 14. A person being asked his age, replied, I am 3 times the age of my son, my son is g of his mother's age, and my wife was 24 what was his age? 15. How much water must be mixed with rum at 18s. per gallon, so as to make 100 gallons worth 15s, per gallon? EQUATIONS CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 507. When there are several unknown quantities, there will be as many separate equations; from these an equation is deduced containing but one of the unknown quantities. This equation is then to be resolved by the foregoing method. 508. There are three methods by which equations, containing several unknown quantities, may be reduced to equations containing but one unknown quantity. METHOD 1. 509. Find the value of one unknown quantity in each of the equations, by placing it separately on one side of the equation, and the several other quantities on the other. From this deduce the value of another unknown quantity, and so on, until an equation is obtained containing but one unknown quantity, which must be resolved as before. 5y 40, what is the value of r If x+y=16, and 5x · and of y? If x + y = 16, then x = 16 then x = 10+5y; therefore, 16 — y — 40 + 5y =8+y, for if two equal quantities be each equal to a third quantity, they are equal to one another; then if 16- y = 8 + y, 16 82y; hence y=4, and x = 12. METHOD 2. 510. Find the value of one unknown quantity in the simplest of the equations, and use this value to find the value of the other unknown quantities. Thus, taking the example given above, viz., x+y=16, and 5r 5y 40; then + y = 16, therefore x = 16 · y; substituting 16 ceed to resolve the other equation, 5 (16 — y) 5y5y40. By transposition 80 Hence y 4, and 16 — y = 12. METHOD 3. y for x, we pro5y=40 or 80 40 40 10y 511. Multiply or divide the equations by such quantities as will make the coefficients or powers of one of the unknown quantities the same in all the equations; then if the signs of the equal terms be like, subtract the equations, but add them, if the signs be unlike. From this process a new equation will result, from which one unknown quantity will be expunged, which must be treated as before, if necessary. We will take the same example as before to illustrate this process : Therefore 12, and as x + y = 16, then y = 4. Obs.-The latter of these methods is the most simple, as well as the most expeditious; but the two former are more scientific, and are useful in strengthening and improving the reasoning faculties. 512. EXAMPLES. 1. What is the value of x + y = 16 and x2 By method third. ≈+y=25 2y= 8 Hence y 4 and x = 12. 2. What is the value of the equations x + y = 30, y + ≈ 25, and x + x = 35? x+x=35 x+y=30 -X Therefore y = 10 513. PROMISCUOUS QUESTIONS IN EQUATIONS. 1. A person with an assistant can perform a piece of work in eight days which he himself could do in 12 days; in what time could the assistant do it? 2. A horse dealer sold a horse for £30, and gained 20 per cent. by the transaction; what did he buy him for? 3. A person had a certain sum of money given him to distribute among the poor of his parish, he found that he wanted 10s. to enable him to give them 5s. a-piece; and that if he gave them 4s. each he would have 5s. left: how much money had he to divide; how inany poor persons were there in the parish; and how much ought they to have a-piece? 4. Two persons set out at the same time to meet one another from two places distant from each other 380 miles ; one travels at the rate of 30 miles per day, the other 25 miles ; in what time will they meet, and how many miles will each travel? |