SIMPLE EQUATIONS.] 



MATHEMATICS. ALGEBRA. 



457 



the left side of which is occupied wholly with un- 

 known, and the right side wholly with known quantities. 

 Reducing now each side to a single term, the equation is 

 2z = 10 ; and dividing by 2, there results x = 5. The 

 steps of the solution, as you would be expected to solve 

 the equation, would stand as follows : 



Given equation, 5x 8 = 3,-c + 2 

 By transposition, 6z 3x = 2 -j- 8 

 Collecting the terms, 2x = 10 



Dividing by 2, x 5 



2 Given the equation 2x 9 = 31 3x to find the 

 value of x. 



By transposing, 2x + 3x = 31 + 9 

 Collecting, 5x = 40 



Dividing by 5, x = 8 



3. Given the equation?? 



o 



value of x. 



Multiplying by 3, to clear the first fraction, 



2x + ^ 66 = 



Multiplying by 4, to clear the remaining fraction, 



8x + 3x 264 = 

 Transposing, 8x + 3x = 264 

 Collecting, 11* = 264 



Dividing by 1 1, z = 24 



4. Given ^ + j + = 1, to find the value of x. 



3?* SB 



Multiplying by 3, x + ~j~ + "~ = 3 



fL 22 = to find the 

 4 



by4, 



by 5, 20* + 15* + 12* 

 * .-. 47x = 60, .-. 



60 



CO 



'47 



EXAMPLES FOK EXERCISE. 



find the value of x in each of the following equa- 

 tions : 



NOTE. Since any quantity may be removed from one side of an equation 

 to the oilier, provided its styn be changed, it follows that if all the 

 signs on one side, and all the sifrns on the other be changed, the 

 equality of the two side* will not be disturbed; for this changing of 

 all the signs is merely the same as trantpoiuig the two sides. 



2. 

 3. 

 4. 

 5. 

 6. 

 7. 

 8. 

 9. 

 10. 



5x 3 = 3.r-t-5 

 6*4-2 = 4*4-8 

 3x l=* + 7. 

 8x 13 = 4* 1 

 2(*+3) = 20 

 3(x 4) = 6 

 4(* l) = *4-5 



3(2* 3) = 4(* 2) 



ii. 4--=; 



12. 

 13. 

 14. 



13. 

 1C. 



+*=3l-f 

 2 ' 3 5 



*+*+?= 1 



3^ 4^0 



2.r (48 *) = *4-12 



17. 



i 



-2-- 



19. ?4-5--l=0 

 2 ' 3 5 



20. +5-*- l _=o 



2 T 3 4 2 



21. ll-5f=f-?f 



823 



22. 4x 2(3 *) = 



23. a,_<=!* = ?? 2 



5 S 



25. 3f=L 3 + 1 J^L =0 



26 C r ~~ 1 -f^?- 3 * 

 2 3 "~T~ 



QUESTIONS TO BE SOLVED BY SIMPLE EQUATIONS. 



Every question, whether in arithmetic or algebra, has 

 for its object the discovery of some unknown value by 

 means of the conditions, stated in the question, which 

 connect it with known values. In algebra, the first thin,' 

 to be done is to express these conditions in the form of 

 an equation, some letter, as x, being made to stand for 

 the unknown quantity; the solution of the equation is 



The sign . . , stands for the word thcrrfort ; on account of its con- 

 venience it will be frequently uicd hereafter. 

 VOL. 1. 



the solution of the question. If there are several unknown 

 quantities to be determined, as many distinct equations 

 will in general be necessary ; at present those questions 

 only will be considered that furnish a single equation 

 with one unknown quantity. The method of translating 

 such questions into equations will be better learnt by the 

 study of a few examples than by verbal directions. We 

 shall therefore here give some instances of the modes of 

 proceeding. 



1. There are two numbers of which the difference is 8 

 and the sum 38 : what are the numbers ? 



Let x stand for the smaller of the two mimbers : then 

 x -f- 8 must represent the greater : and since by the ques- 

 tion the sum of the two is 38, we have the equation, 



x + x + 8 = 38. 



Or transposing the 8, 2* = 30, .*. x = 15, the smaller 

 number ; and as x + 8 is the greater, /. 15 + 8 = 23, is 

 the greater. 



You see that these numbers satisfy the conditions of 

 the question ; for 23 15 = 8, and 23 + 15 = 38. 



2. From two places, 160 miles apart, two persons, A 

 and B, set out at the same time to meet each other. A 

 travels 18 miles a-day, and B 22 miles : in how many 

 days will they meet ? 



Suppose they'meet in x days : then A will have travelled 

 18.r miles, and B, 22.r miles : the sum of these distances 

 is, by the question, 160 miles : hence we have the equa- 

 tion, 



l&r 4- 22x = 160 ; 

 that is, 40* = 160, .'. x = 4, the number of days. 



3. A father is three times as old as his son ; but five 

 years ago he was four times as old : what are their ages 

 now? 



Let the present age of the son be x years : then, by 

 the question, that of the father is 3x years. Also five 

 years ago the age of the son must have been x 5, and 

 that of the father Sx 5. The question tells us that 

 this latter number is four times the former : that is, 



Sx 5 = 4(> 5); 



or, removing the vinculum, 3x 5 = 4z 20 ; /. trans- 

 posing the 20 to the left, and the 3x to the right, 



20 5 = 4x Sx, .'. 15 = x; 



that is, the son's age is 15 years ; and 3.r, the age of the 

 father, is 45. 



4. A vessel holding 120 gallons is partly filled by a 

 spout which delivers 14 gallons in a minute : this is then 

 turned off, and a second spout, delivering 9 gallons in a 

 minute, completes the filling of the vessel. How long 

 did each spout run, the time occupied by both in filling 

 the vessel being 10 minutes ? 



Suppose the first ran x minutes : then the second ran 

 10 x minutes. As the first delivers 14 gallons a 

 minute, the quantity delivered by this spout must be 

 14x gallons ; and the quantity delivered by the other, 

 at 9 gallons a minute, must be 9(10 x) gallons ; .*. the 

 whole number of gallons delivered is, 



14x 4- 9 (10 x) = 120, by the question ; 



that is, removing vinculum, 14z-|- 90 9j; = 120 ; 



/. transposing, 14x 9x = 120 90 ; 



collecting, 5x = 30, /. x = 6. 



.'. the first spout ran 6 minutes, and the second 10 6 

 = 4 minutes. 



5. What number is that of which the third part exceeds 

 the fifth part by 48 ? 



Let x be the number : then by the question 



Multiplying by 3, *-=144 



P 



by 5, 5* 3.r = 720! 



that is, 2x 720, .'. x = SCO, the number required. 

 The third part of this is 120 ; the fifth part is 72 ; and 

 the difference between these parts is 48. 



0. A vessel can be filled from a tap in 3 hours, and 

 from a second tap in 5 hours : in what time will it be 

 filled if both taps run together ? 

 Lr.t the number of hours be x : then tho part of the 



vessel filled in one hour will be the - part. 



x 



3 v 







