ALGEBRA 



194 



ALGEBRA 



(1) 



3a+46=34 

 ................... 6+36=33 



(3) Multiplying (1) by 2 ........ 6a+86=68 



(4) Bringing down (2) ........... 6a+36=33 



(5) Subtracting .................. 56=35 



(6) .............................. b= 7 



ng found the numerical value of b, it is 

 an easy matter to apply that value in cither 

 of the original equations; in other words, to 

 substitute in equation (1) or equation (2) the 

 value of b. 



Since we know that b equals 7 we use the 

 value of 46, or 28, in the first equation, so that 

 our statement now reads: 3a+28=34. In or- 

 der to get both known quantities on the same 

 side of the equality sign we must transpose the 

 28. Then we have, 3a=34 28; 3a=6, and 

 0=2. The formal statement for the comple- 

 tion of the problem is: 



7 Applying the value of b in (1) 3a+28=34 



(8) Transposing ................ 3a=34 28 



(9) Then ....................... 3a=6 



(10) And ........................ a=2 



(11) Proof .......... 3a, or 6, +46, or 28, =34 



For practice, solve the problems given below, 



using either addition or subtraction to elimi- 

 nate unknowns. The beginner will find it 

 helpful to write out each solution fully, put- 

 ting down the steps in order and thus making 

 himself familiar with the principles involved. 



2. Solve | 



3. Solve 



y=lQ 

 2y=2Q 

 x+ 5?/=34 

 [4H- 3?/=51 

 Elimination by Substitution. By this is 

 meant the process of clearing an equation of 

 one of its unknown terms by substituting in 

 either equation the value of one of its unknown 

 terms, as in the following solution: 



(2a-f56=31 

 Solve {3a+4y=29 

 We will first transpose 2a in the first equa- 



tion and thus get a statement for the value 

 of 6. 



Transposing 2a. \\v have 56=31 2a, and 6= 

 31 2a 



Now wo write the second equation of the 

 problem, placing the new value of 6 in it. We 

 thus have: 



. (31 2a) 



3a+4 =29 



o 



Before this equation can be simplified it must 

 be cleared of fractions, as follows: 



(31 2o) 

 3a+4 =29 



(124 8a) 

 3aH =29 



15a+124 8a=145. 



Transposing the known quantities to the 

 right of the equality sign, we have: 



15a 8a=145 124, or 7a=21. 

 The following statements show the entire 

 process step by step: 



(i) 



(2) 



2o + 5b = 31 

 3a + 4& = 29 

 312o 



(3) Transposing 2a in (1) 



and dividing by 5 u 5 



(4) Substituting the value (31 2o) 



of b in (2) 3o+4 = =29 



5 



(5) Clearing of fractions... 15o + 124 80 = 145 



(6) Transposing 15a 8a = 145 124 



(7) . a = 3 



( 8 ) Substituting the value 

 of a in (3) 



31-6 

 " 5~ 



(9) 



For practice, solve the following problems, 

 using the method of eliminating by substitu- 

 tion: 



C 



1. Solvej 



5x3y= 8 



2. Solve 



3. Solve 



Problems Involving Three Unknown Quantities 



Problems involving three unknown quanti- 

 ties, though a little more complicated, present 

 no special difficulties, for they may be solved 

 by applying the rules for elimination to two 

 of the given equations, and when the values 

 of two unknown quantities are found, these may 



be substituted in connection with the third 

 unknown quantity. The full solution of such 

 a problem is given below: 



( x+ y+ 2=10 

 SolveJ 3z+22/+4z=33 

 = 9 



