ALGEBRA 



ISO 



ALGEBRA 



with the principle it will be seen that a as well 

 as 5 is a coefficient that 5 may be taken a 

 times. So, really, 5 is the coefficient of a, and 

 a is the coefficient of 5. To apply the principle 

 further, in a(x+j/), a is the coefficient of 

 OH- y) and (x+y) is the coefficient of a. 



Signs of Parentheses. In the above para- 

 graph we have learned that if several numbers 

 or letters are to be treated as a single expres- 

 sion they are joined together by being enclosed 

 in parentheses. There are two rules laid down 

 for guidance in treating such aggregations. The 

 first is usually stated in this form: 



If an expression within parentheses Is pre- 

 ceded by the sign + , the parentheses can be 

 removed without making any change in the signs 

 of the expression, and without altering values. 



It is a simple matter to prove this to be 

 true. Let us do it in this way : 



If a man has 40 dollars and later collects 8 

 dollars and then 2 dollars, it is immaterial 

 whether he adds the 8 dollars to his 40 dollars, 

 and afterwards adds the 2 dollars, or whether 

 he adds to his 40 dollars the sum of 8 dollars 

 and 2 dollars. 



The first process may be represented thus: 

 40 + 8 + 2. 



The second process may be represented thus: 

 40+(8 + 2). 



Hence, 40+ (8 + 2) =40 + 8 + 2. 



Again, if the same man has 40 dollars and 

 later collects 8 dollars and pays a debt of 2 

 dollars, it Is immaterial Whether the 8 dollars 

 be added to the 40 and the debt be paid out 

 of the sum, or whether the 2 dollars be paid 

 out of the 8 dollars and the remainder be added 

 to the 40 dollars. 



In the first case the process is represented by 

 40 + 82. 



In the second it is represented by 40 +(8 2). 



Hence, 40 + ( 82 ) = 40 + 82. 



Prove that you understand the principle by 

 simplifying the following: 



5-M8 4) =5+8-4. 

 The simple form is 9=9. 



4+ (8-2) + (6+1) =4+ 8-2+64-1. 

 There should not be the slightest difficulty in 

 toying the above rule. The second needs a 

 little deeper study: 



If an expression within parentheses Is pre- 

 ceded by the sign , the parentheses can be 

 removed, provided the sign before ea< h t 

 within the parentheses In changed, the sign + 

 to , and the sign to +. 



r>t illn-t ration, now that we are famil- 

 iar with it, may be used with one variation, in 

 explaining this second rul : 



If a man has 40 dollars and has two bills to 

 pay, one of 8 dollars and one of 2 dollars, It is 



immaterial whether he takes the 8 dollars and 

 2 dollars one after the other, or whether he 

 takes the 8 dollars and the 2 dollars at one 

 time from the 40 dollars. 



We may represent the first process by 40 8 2. 



We may represent the second by 40 (8 + 2). 



Hence, 40 (8 + 2) =40 8 2. 



If this man has his 40 dollars in the form 

 of five-dollar bills, and has a debt of 8 dollars 

 to pay, he can do so by giving two bills (10 

 dollars) and receiving 2 dollars in return. 



We may represent this process by 40 10 + 2. 



If the bill paid Is 8 dollars, that is. (102) 

 dollars, the number of dollars remaining may 

 be represented by 40 (10 2). 



Hence, 40 (10 2) =40 10 + 2. 



To make sure your understanding of the 

 above, solve the following: 



6 (4 2) = ? 



9 (4 + 3) = ? 

 (6 2) (5 2) = ? 

 12 (8 3 2) = ? 

 15__(6 2 + 3) = ? 

 (14a-2a)-(6o-2a) = ? 



Numerical Values of Letters. Referring again 

 to the term coefficient, we recall that a coeffi- 

 cient is a multiplier. Thus, in the expression 

 8a, a, no matter what its value may be, is to 

 be taken 8 times. It follows then that if a=3, 

 the expression 2a+3a=6+9, or 15; or, 2a+3a 

 =5a, or 15. When no coefficient is expressed 

 it is understood to be 1 ; a=la, 6=16, etc. 

 Apply these facts in the solution of the follow- 

 ing problems. In a few instances results are 

 stated to make the mastery of the principles 

 easier. 



If a=4, 6=3, c=2, find the value of: 



1. 80 36c. (Am., 14; in this case 6 and c 

 are to be multiplied together, and 3 is their 

 coefficient). 



2. 4ac+5a. 



3. 2(a 6+c). 



4. 6+3(a c). Ana., 9. 



5. 46 2(a+c). 



6. 8c 6(a 6). 



The last problem is here solved step by step. 

 Compare with your own solutions and see if 

 your methods are correct: 



(1) 8c 6(0 6). 



(2) Removing parentheses, we have 8c 06+ 

 6 s ; & times b is not 26, but 6 1 , because 6 is 

 multiplied by itself. 



(3) Assigning values to the letters, 

 1612+9. 



Adding the terms with plus signs, and sub- 

 tracting from their sum the term having a 

 minus HUM. 



(4) 25-12=13. 



