156 



THE POPULAR EDUCATOR. 



the one side of the equation, and all the known quantities to the 

 other, taking care to change the signs of the terms transposed, and 

 incorporate the terms that are alike. 



4. Remove the co-efficient of tlie unknown, quantity, by dividing 

 all the terms in, the equation by it ; the result will be the solution 

 required. 



PROOF. Substitute the value of the unknown quantity for the 

 letter which stands for it in the equation ; and if the number 

 satisfies the conditions of the question, it is the answer sought. 



PROBLEM 1. A man being asked how much he gave for his 

 watch, replied : If you multiply the price by 4, to the product 

 add 70, and from this sum subtract 50, the remainder will be 

 equal to 220 pounds. 



In order to solve this question, we must first translate the 

 conditions of the problem into such an algebraic expression as 

 will form an equation. 



Let x be the price of the watch. 



This price is to be multiplied by 4, which makes 4o; ; to tho 

 product 70 is to bo added, making 4# + 70 ; from this, 50 is to 

 be subtracted, making 4^+70 50. 



Here we have a number of the conditions, expressed in 

 algebraic terms ; but we have as yet no equation. We must 

 observe, then, that by the last condition of the problem, the 

 preceding terms are said to be equal to 220. 



We have, therefore, this equation, 4 + 70 50 = 220 ; 

 which reduced, gives x = 50. Ans. 



Here the value of x is found to be 50 pounds, which is the 

 price of the watch. 



PROOF. The original equation is 4a; + 70 50 = 220 ; sub- 

 stituting 50 for x, it becomes 4 X 50 + 70 50 = 220 ; that 

 is, 220 = 220. 



PROBLEM 2. What number is that to which, if its half be 

 added, and from the sum 20 be subtracted, the remainder will 

 be a fourth of the number itself ? 



In stating questions of this kind, where fractions are con- 

 cerned, it should be recollected that |x is the same as - ; that 

 \x= -, etc. 



Let x be tho number required. 



Then by tho conditions, we have x + 20 = -, and re- 



2 4 



ducing .the equation, we have x = 16. Ans. 



PROOF. Thus 16 + 20 = . 



2 4 



PROBLEM B. A father divides his estate among his three sons 

 in such a manner, that the first has .1,000 less than tho 

 whole ; the second has .800 less than one-third of the whole ; 

 the third has .600 less than one-fourth of the whole. What is 

 the value of the estate ? Ans. <4,114f. 



PROBLEM 4. Divide 48 into two such parts, that if the lesa be 

 divided by 4, and the greater by 6, the sum of the quotients 

 will be 9. 



Let x be the smaller part ; then 48 x is the greater part ; and, 



by the conditions of the problem, we have - -f- ^^ = 9. 



Whence x 12; therefore, 12 is the less part, and 36 the 

 greater part. 



171. Letters may be employed to express the known quantities 

 in an equation, as well as the unknown. A particular value is 

 assigned to the letters, when they are introduced into the calcu- 

 lation ; and at its close, the numbers are restored. 



EXAMPLE. If to a certain number 720 be added, and the 

 Bum be divided by 125, the quotient will be equal to 7392 

 divided by 462. What is the number ? 



Let x be tho number required; and let a = 720, 5 = 125, 

 t = 7392, and h = 462. 



Then, by the conditions of the problem, we have 1_ = _ ; 



o h 



and reducing, we have x = 



bd ah 



T, . . ., , , (125x7392)-(720x462) 

 .Restoring the numbers, we have x = ' v 



= 1280. 462 . 



EXERCISE 29. MISCELLANEOUS PROBLEMS IN SIMPLE 



EQUATIONS. 



1. Divide 11 into two parts, such that the sum of twice the 

 first and half the second may be 16. 



2. Divide 39 into four parts, such that if the first be increased 

 by 1, the second diminished by 2, the third multiplied by 3, and 

 the fourth divided by 4, the results may be all equal. 



3. If a certain number is divided by 12, the quotient, 

 dividend, and divisor, added together, will amount to 64. What 

 is the number ? 



4. An estate is divided among four children, in such a manner 

 that the first has .200 more than | of the whole, the second 

 has .340 more than | of the whole, the third has .300 more 

 than > of the whole, the fourth has .400 more than A of the 

 whole. What is the value of the estate ? 



5. What is that number which is as much less than 500, as a 

 fifth part of it is greater than 40 ? 



6. There are two numbers whoso difference is 40, and whick 

 are to each other as 6 to 5. What are the numbers ? 



7. Suppose two coaches to start at tho same hour, one from 

 London for Glasgow, and the other from Glasgow for London, 

 the former travelling 10 and the latter 9 miles per hour. 

 Where will they meet, the distance between the two cities being 

 400 miles ? 



8. Suppose everything to be as in the last question, except 

 that the coach from Glasgow starts two hours earlier than the 

 other. Where will they meet ? 



9. A dealer purchases 60 yards of cloth for .30 ; and by 

 selling one part of it at 12s., another, twice as great, at 14s., 

 and the rest at 10s. per yard, he gains .8. How many yards 

 were in the several lots ? 



10. Suppose two dealers each annually to double his capital, 

 except an expenditure of ,100 ; and that at the end of three 

 years the capital of one is found to be doubled, while the other 

 has only half what ho had at first. How much had each to com- 

 mence with ? 



11. If a person eaah year double his capital, except an expen- 

 diture of .300 the first year, .400 the next year, and .500 the 

 third, and at the end of three years be found to be worth .5,500, 

 what was his original capital ? 



12. A father's ago is now treble of his son's, while five years 

 ago it was quadruple. What are their present ages ? 



13. Divide .1,000 between A, B, and C, giving A 100 more, 

 and B 50 less, than C. 



14. A spirit merchant finds that if he add 10 gallons to a cask 

 of brandy, the mixture will be worth 21s. per gallon ; but that 

 if he had ten gallons more, the value will be reduced to 18s. 

 How many gallons were in the cask ? 



15. Find a number, such that if it be divided successively by 

 2, 3, 4, 5, 6, 7, 8, 9, and 10, half the sum of the first four 

 quotients increased by 20 shall be equal to the sum of the 

 remaining five. 



16. Find two numbers differing by 6, and such that three 

 times the less may exceed twice tho greater by 7. 



17. Find a number, such that if it be increased successively 

 by 1, 2, and 3, the sum of one-half of the first result and one- 

 third of the second shall exceed one-fourth of the third by 8. 



LESSONS IN ITALIAN. XIV. 



EXERCISES FOB PRACTICE. 



WE resume in this lesson our series of exercises which will 

 afford the student sufficient practice in translating simple sen- 

 tences in Italian into English, and turning English into Italian. 

 The copious vocabularies will afford the learner a useful oppor- 

 tunity of storing his mind and memory with Italian words. 

 VOCABULARY. 



L' o-ro-16-gio, the watcL 



or clock. 

 Ma, but. 

 M6l-to, very. 

 Per, for. 

 Per-du-to, lost. 

 Tro-vd-to, found. 



EXERCISE 8. COLLOQUIAL. 

 1. Mi-o pa-dre e buo-no ; e-gli ha an-che un bu6n fra-tel-lo. 



2. Mi'-a ma-dre e bu6-na ; el-la ha an-che u-na buo-na so-rel-la, 



3. Ab-bia-mo ve-du-to v6-stro zi-o ; e-gli ha com-pra-to un gran 

 h'-bro. 4. A-ve-te vdi ve-du-to il n6-stro giar-di-no P^es-so e 

 mol-to gran-do. 5. H6 com-pra-to u-na pen-na ; es-sa c mdl-to 



