ALGEBRA 



193 



ALGEBRA 



an equation and still preserve its equality, we 

 may write: 



10a+22=322. 

 This is equivalent to 10a=32 2. 



Simplifying, 10a=30. 



Dividing both sides by 10, a =3. 



These operations \ are fundamental in what 

 is known as transposition. When we change 

 the form 10a-f 2=32 to the form 10a=32 2, 

 we transpose a known quantity from one side 

 of the equation to the other, and in doing so 

 we change its sign. In the solution of simple 

 equations by transposition we work with two 

 principles: known quantities are placed on one 

 side of the equality sign and unknown quanti- 

 ties on the other; any quantity may be trans- 

 posed from one side to the other if its sign is 

 changed. As we learned above, when we trans- 

 pose a quantity we are really adding it to both 

 sides of the 'equation or subtracting it from 

 both sides. Let us see how these facts are ap- 

 plied in the solution of practical problems: 



1. A ditch 80 feet long is divided into two 

 parts in such a way that one part is three 

 times as long as the other. What is the length 

 of each part? 



As the length of neither part is known, we 

 may represent the number of feet in the shorter 

 part by x. The solution is stated thus: 



Let x = number of feet In shorter part. 

 3x = number of feet in longer part. 

 ar-f-Sx, or 4x, = number of feet in total 



length. 

 4s = 80. 



a? = 20, number of feet in shorter part. 

 3x = 60. number of feet in longer part. 

 Proof: 60 = 3X20. 60 + 20 = 80. 



2. Find a number such that when 14 is added 

 to twice the number the sum will be 64. 



I If we represent the number to be found by x, 

 we know that twice the number must be 2x. 

 Since 14 added to double the number equals 

 64, our next statement must read, 



2x+14=64. 



Applying the rules we have just learned, we 

 have, 



2r=64 14, 



2x=50, 



x=25, required number. 

 Solve the following problems, using the above 

 explanations as your guide: 



1. The sum of two numbers is 60, and the 

 greater is five times the less. What is each? 



2. A man divided 75 dollars between two 

 sons. To A he gave twice as much as to B. 

 How much did each receive? 



3. Four times a certain number is equal to 

 the number Increased by 36. What is the 

 number? 



4. John bought a certain number of apples. 

 Had he bought three times as many he would 

 have had 20 more than the original number. 

 How many did he buy? 



5. An orchard yields 140 bushels of fruit. 

 Hint : Let x = number of bushels of peaches, 



is 15 more than the number of bushels of pears. 

 Find the number of bushels of each. 



Hint : Let x = number of bushels of peaches, 

 and 140 x = the number of bushels of pears. 



6. A farm of 160 acres was divided into three 

 sections. The first was twice the size of the 

 second, and the second three times the size of 

 the third. What was the acreage of each? 



7. A 45-acre farm was divided into three 

 garden plots. The first was half the size of the 

 third and the second half the size of the first 

 and third combined. How many acres In each? 



Problems Dealing with Two Unknown Quantities 



As the next step in our work we take up 

 equations in which two unknown quantities 

 pteur. Though such problems are more com- 

 plex than the ones given above, their solution 

 ,is not difficult if the philosophy of the simple 

 equation is clearly understood. \ 



Usual Methods. In solving problems involv- 

 ing two unknown quantities several methods 

 may be used to eliminate the unknowns. To 

 Himimite an unknown quantity is to find its 

 numerical value; the new value is then placed 

 rial equation as a substitute for the 

 1 unknown quantity. The methods most 

 mnunonly used are elimination by addit 

 subtraction, and elimination by substitution. 

 ! otlu r methods, but they are less fre- 

 quently employed than the ones explained here. 

 13 



By Addition or Subtraction. Let us examine 

 step by step the process of finding the value 

 of two unknown quantities by the first method. 

 This method is usually the simpler and easier 

 of the two. 



Solve 



[3a+46=34 

 !6a+36=33 



It is clear tli.it if the first equation is multi- 

 plied by 2, it will l>e in Mich form that the first 

 term will equal th.- first terra of the second 

 equation. The entire product will read: 6a+ 

 86=68. If we subtract the second equation 

 from the first a* it now stands, we will have a 

 remainder of 56=35. The value of 6 is then 

 easily found. The various steps of the process 

 arc shown in the following statements: 



