314 



THE POPULAB EDUCATOR. 



the spark, the time required for its passage being so short that 

 the light has faded before the wheel has moved through an 

 appreciable space. 



If the points be made nearly to touch, and the thumb 

 pressed down upon them, it will appear to be rendered semi- 

 transparent, and the shock will not be felt. In a similar way 

 the shock may be passed through eggs, oranges, etc., illumi- 

 nating them. Some substances, too, remain luminous for some 

 time after the charge has passed through them. 



If we fix a ball to the plate of an air-pump, and connect it 

 with the exterior of a jar, and then place over it an open re- 

 ceiver with a similar ball and rod passing air-tight through a 

 plate at its upper end, we shall find that the spark will pass a 

 much greater distance. In a long exhausted tube it will some- 

 times appear to dart along like a ball of light. 



Similarly, if we exhaust a large globe having a wire and ball 

 fixed tightly in each end, and connect the upper one B (Fig. 20) 

 with the prime conductor, we shall find that, instead of the elec- 

 tricity passing along in a series of thin sparks, it will spread out 

 and form a large faintly luminous space, having a violet tinge. 

 On again gradually admitting the air by means of the stopcock, 

 this space will grow smaller and smaller, till at last the narrow 

 but bright sparks pass as before. Many similar experiments 

 of great beauty may be performed by means of voltaic elec- 

 tricity, and it is believed that the beautiful phenomena of 

 the aurora boreali's, or northern lights, so frequently seen in 

 high latitudes, may be attributed to a similar cause. 



LESSONS IN ALGEBRA. XXII. 



SIMPLE EQUATIONS. 

 TWO UNKNOWN QUANTITIES. 



ITT our former Lessons on Simple Equations we gave the rules 

 for solving those which contain only one unknown quantity ; 

 and, with the exception of one or two, the whole Centenary of 

 Problems were solved by means of these rules. We proceed 

 now to show how to resolve equations which contain two un- 

 known quantities. 



Cases indeed frequently occur in which two unknown quan- 

 tities are necessarily introduced into the same calculation. 



EXAMPLE. Suppose the following equations are given, 

 viz. : 



(1.) x + y = 14, 

 (2.) x y = 2. 



Here, if y bo transposed in each, they will become 

 (1.) * = 14 - y, 



Now, the first member of each of the equations is x, and the 

 second member of each is equal to x. But according to the 

 axiom that quantities which are respectively equal to another 

 quantity, are equal to each other ; therefore we have 

 2 -)- y = 14 - y ; whence y = 6. 



Lastly, by substituting the value of y in the 1st equation, we 

 have x + 6 = 14 ; and as = 8. Therefore, 8 and 6 are the values 

 of x and y. 



In solving the preceding problem, it will be observed that 

 we first found the value of the unknown quantity x in each 

 equation ; and then, by making one of the expressions denoting 

 the value of x equal to the other, we formed a new equation, 

 which contained only the other unknown quantity y. This pro- 

 cess is called extermination or elimination. 



In the resolution of equations there are three methods of 

 extermination, viz., by comparison, by substitution, and by 

 addition and subtraction. 



CASE I. To exterminate one of the two unknown quantities 

 by comparison. 



RULE. Find the value of one of the unknown quantities in 

 each of the equations, and form a new equation by making one of 

 these values equal to the other. Find the value of the unknown 

 quantity in this equation, by the rules formerly given. Then 

 substitute this value of the one unknown quantity in either of the 

 other equations, and resolving it by the same rules, the other 

 unknown quantity will be found. 



EXAMPLE. Given x -f y 36, and x y = 12 ; to find the 

 values of and y. 



Transposing y in the 1st equation, gives x = 36 - y. 

 Transposing y in the 2nd equation, x = 1 2 + y. 



Making these values of x equal, ,, 12 + y = 36 y. 

 Transposing, etc., ,, 'y=12. 



Substituting the value ofy, = 12-fl2 = 24. 



Hence, 24 and 12 are the values required. 



EXERCISE 36. 



1. Given 2* + 3y = 28, and 3* + 2y = 27 ; to find the values of x and y. 



2. Given 4* + y = 43, and 5* + 2i/ = 56 ; to find the values of x and ij. 



3. Given 4* 2y = 16, and Gx = 9y ; to find the values of x and y. 



4. Given 4x 2y = 20, and 4x + 2y = 100 ; to find the values of a 



and i/. 



5. Given 5* + 8 = 7y, and 5y + 32 = 7 ; to find the values of x and &. 



EXAMPLE (1). To find two numbers such that their suns 

 shall be 24 ; and the greater shall be equal to five times the less. 

 Here, let at be the greater ; and y the less. 



Then, * -f y = 24, 



And x = 5y. 



Whence, 5y -j- y = Gy = 24, 



And y = 4 ; 



Therefore, x = 20. Ans. 20 and 4. 



EXAMPLE (2). Find two quantities whose sum ia equal to fc-j 

 and the difference of whose squares is equal to d. 

 Let x and y be the two quantities. 



Then x +y = h) 

 , j * i j f per question. 

 And z 2 y 2 = a \ * 



From the first equation we have, by transposition, 



x = h y, 

 And, by squaring both sides, we have, 



x 3 = h* 2hy + y 2 . 

 From the second equation, we have, by transposition, 



* = 2 + d. 

 Now, by equating the two values of x 1 , we have, 



y2 + d = h 2 2hy + y- ; 



And, by transposition and cancelling, we have, 

 2hy = h~ d; 



, !<? d 



Whence, y = - -- 



Therefore, * = h- *-- * = J *+J. 

 2h 2h 



EXAMPLE (3). Given ox + by = h t and x + y = d ; to find 

 the values of x and y. 



Here, from the first equation, we have, by transposition, 

 ox = h by, 



Again, from the second equation, we have, by transposition, 

 = d y, 



Whence, ^ d y ; 

 a 



Or, h by ad ay, 



And ay by = ad h. 



From this equation, by separating the left-hand member in 

 factors, we have 



(a 6) y = ad h ; 

 ad h 



Whence, 



y = 



Consequently, x = d 



ad h h bd 



a b a b 



The rule given above may be generally applied for the exter- 

 mination of unknown quantities. Bat there are cases in which 

 other methods will be found more expeditious. 



EXAMPLE (4). Given x = hy, and ax + bx = ?/*; to find the 

 values of x and y. 



As in the first of these equations x is equal to 7iy, we may in 

 the second equation substitute this value of x for x itself. The 

 second equation will then become, ahy -f- bhy = y 2 . 



The equality of the two sides is not affected by this altera- 

 tion, because we only change one quantity x for another which 

 is equal to it. By this means we obtain an equation which 



