Solution of Equations in x and y 224. If 4 bananas and 9 oranges cost 35, and 4 bananas and 6 oranges cost 264, and it is required to find the cost of 1 of each, we may simplify the problem thus: By thus eliminating the cost of the bananas, we have obtained a relation, (3), more simple than either of the two given relations, (1) and (2), for it involves only one unknown cost. Or, let x be the number of cents 1 banana costs, and y the number of cents 1 orange costs. Then 4 bananas will cost 4 x cents, 9 oranges 9 y cents, etc. = Since y 3, 9 y in the first equation is equal to 27. To test the answers we substitute 2 for x and 3 for y in the Therefore the values obtained for x and y satisfy both equations, and the answers are correct. This method of elimination is called elimination by subtraction. THIRD PROG. AR. - -9 In eliminating the x's on page 129, equal numbers, 4x + 6 y and 26, are subtracted from both members of (1). Therefore the results are equal, giving a true equation, 3y = 9. If equals are subtracted from equals, the results are equal. 225. How must the equations 2x+3y= 16 and 5x-3y=19 be combined to eliminate the y's? This method of elimination is called elimination by addition. If equals are added to equals, the results are equal. WRITTEN EXERCISES 226. 1. If 2 x + 3y = 18 and 2x + y = 10, what is the value of each unknown number? NOTES.-1. The sign .. means "therefore." 2. The value of y may be substituted in either of the given equations. 14. If 2x+3y= 16 and 5x + 4y = 33, find x and y. We may eliminate either x or y. If we choose to eliminate x, we must first prepare the equations, so that x may have the same coefficient in each. Multiplying both members of (1) by 5, and both members of (2) by 2, Test. NOTE. These values give 10 + 6 = 16 and 25+ 8 = 33 in (1) and (2). To eliminate y instead of x, proceed as follows: Subtracting the upper equation from the lower, thus avoiding negative coefficients, 7x=35; .. x = 5 Substituting 5 for x in (1), 10+ 3y = 16; y = 2 Equations in this book are intended to give aid in solving the more difficult problems of arithmetic, and are not given as exercises for their own sake. To keep within the limits of arithmetic such arithmetical problems or equations as, "Find two numbers whose sum is 10 and whose difference is 2," } or [ x + y = 10 should be solved by eliminating by addition or subtraction the unknown number found in the negative term or terms, if there SUGGESTION. Multiplying the members of each equation by the 1. c. m. of the denominators in that equation will clear the equation of fractions. members of the first equation by 6, then, changes it to 9 x Multiplying the Find two numbers related to each other as follows: 32. Sum 14; difference = 8. = = = 34; 20. 33. Sum of 2 times the first and 3 times the second sum of 2 times the first and 5 times the second = 50. 34. Sum 18; sum of the first and 2 times the second 35. A grocer sold 2 boxes of raspberries and 3 of cherries to one customer for 544, and 3 boxes of raspberries and 2 of cherries to another for 564. Find the price of each per box. 36. A druggist wishes to put 500 grains of quinine into 3grain and 2-grain capsules. He has 220 capsules. How many capsules of each size can he fill? 37. On the Fourth of July, 850 glasses of soda water were sold at a fountain, some at 5 each, the others at 10 each. The receipts were $55. How many were sold at each price? 38. A fruit dealer bought 36 pineapples for $2.50. He sold some at 12 each and the rest at 10 each, thereby gaining $1.50. How many did he sell at each price? 39. An errand boy went to the bank to deposit some bills for his employer. Some of them were 1-dollar bills, and the rest 2-dollar bills. The number of bills was 38 and their value Find the number of each. was $50. |