ALGEBRA 



192 



ALGEBRA 



Divide 

 Solution : 



Ox 3 !/* 2 6x 2 y 3 



In long division, for convenience in multiply- 

 :s customary to write the divisor at the 

 ..f the dividend. The following is an 

 ptable form: 



12a 2 +18ab+66 2 I 4a+2b 



12a 2 + 6b 



3a+3b 



12ab+6b 2 

 12ab+6b 2 



We find by inspection that 4a, the first term 

 of the divisor, is contained in 12a 2 , the first 

 term of the dividend, 3a times. Multiplying 

 this partial quotient 3a by the entire divisor, 

 placing the product under the first two terms 

 of the dividend and subtracting, we obtain 

 I2ab. We bring down and add to this re- 

 mainder the next unused term (6b 2 ) in the 

 dividend. By inspection we find that the first 

 term of the divisor is contained in the first 

 term of the new dividend 3b times. We mul- 

 tiply tho entire divisor by 3b and obtain I2ab 

 +6b 2 . Subtracting this product from the new 

 dividend, we obtain no remainder and know 

 that our division is complete. 

 The signs in the above problem are all plus. 

 Note the solution of the following problem, 

 in which minus signs occur: 



a 2 2ab+b 2 \ ab 



a 2 ab a b 



-ab+b 2 

 ab+b 2 



Note that whon we divided the first term in 

 the now dividend, ab, by the first term in 

 the divisor, a, we obtained as a quotient b. 

 Whenever a negative term is divided by a posi- 

 tivn term, or a positive by a negative, the sign 

 of the quotient will be minus. But a minus 

 term divided by a minus term gives a positive 

 quotient. Briefly stated, like signs produce 

 plus, and unlike signs produce minus quanti- 

 ties. 



Solve for practice the following: 



1. Divide Oz 2 18xy+9y 2 by 3z 3y. 



2. Divide a 2 12a-f 35 by a 5. 



3. Divide 3x 4 Wx*y+22x*y 2 22xy*+15y* by 

 x 2 2xy+3y 2 . 



4. Divide a 5 2a 4 4a 3 +19a 2 31a+15 by a 3 

 7a+5. 



Simple Equations. Two or more terms con- 

 nected by the sign of equality (=) form what 

 is known as an equation. The principles upon 

 which the solutions of equations are based may 

 be readily understood by using the familiar 



THE BALANCE SCALE 



balance scale as an illustration. Suppose we 

 have such a scale as is shown in the accom- 

 panying picture. In one pan we place a ten- 

 pound weight; in the other we place a six- 

 pound and a four-pound weight. The first 

 weight, we know, balances the other two, and 

 this fact may be indicated by the following 

 statement : 



10=6+4. 



Suppose we add 5 pounds to each pan. Then 

 our statement is modified to read, 

 10+5=6+4+5. 



If we now remove 3 pounds from each pan 

 we have : 



10+53=6+4+53. 



From these statements, or equations, we may 

 see that the following principles are true : 



1. The same quantity may be added to both 

 sides of an equation, or be subtracted from both 

 sides, without changing the value of the equa- 

 tion. 



2. We may multiply or divide each side of an 

 equation by the same quantity without changing 

 the value of the equation. 



By the application of these rules we may 

 find the numerical values of unknown quanti- 

 ties. In the equation 10a+2=32 we have 

 stated that 32 is 2 more than 10a, or that to 

 10a we must add 2 to equal 32. If we wish 

 to ascertain the number to which 10a is equal 

 we must subtract 2 from 32. Since we may 

 subtract the same number from both sides of 



