International projectionist (July-Dec 1934)

October 1934 INTERNATIONAL PROJECTIONIST Answers to Problems, Lesson X MATHEMATICS FOR THE PROJECTIONIST WHILE it cannot be said that the number of answers received to the final installment helped to close out the series on mathematics in any blaze of glory, it certainly can be said that the quality of answers received was far above the average. A term like "simultaneous linear equation", if hurled at projectionists generally a year or so ago, would have meant but little ; yet now pop up any number of men who not only understand the term but actually work out such problems. As for those who continue to shy at even a simple mathematical problem — well, they simply must plod along and be barred from appreciating enormous quantities of good stuff that just can't be properly explained without a few equations; not to mention the fact that the series now closed should serve to impress upon them how precious little they know, how much there is to learn. As was to be expected, practically the same group of old-timers came through once more with fine sets of answers. Among these were Dale Danielson, Joe Uebelhoer, one of the most painstaking students of the group; Harold F. Miller, a newcomer from Chicago who knocks over the final group in a manner which suggests that he knows what it is all about; Frank Dudiak, who now mixes projection with courses at West Virginia U; Howard Williams, Edward Regula and Raymond Mathewson, who put New Britain, Conn., on the map with a joint set of answers ; Edward Burke, who didn't quite reach 100% but who made a swell job of the series generally; Sam Barr and Mason Sperling. Also, Joe Barnett, with another clean, crisp effort; G. L. Cummings, John Fegan, Ralph Welling, Clifford Newton, and Otis Clarke. Not a few others who plugged along right through the series — ■ their only reward the satisfaction of seeing each month how wrong they were — ■ would receive mention herein if space permitted. These participants probably got far more out of the series than did those to whom the problems came more or less naturally. Appended hereto are the answers to the problems, together with a graphical solution of Problem 5: 1. Solve the following by the method ot substitution: 5x + 6y = 27 x+7y=17 Answer Solving: x=17— 7y Substituting this value for (1): x in equation 5(17— 7y)+6y=27 85— 35y+6y =27 — 29y=27— 85 — 29y=— 58 — y=— 2, or y=2 5x+6y=27 5x+12=27 5x=27— 12 5x=15, or x=3 2. Solve the following by addition and subtraction: 2x— 3y=ll 3x+y=22 Answer Multiplying equation (2) by 3: 9x+3y=66 (3) 2x— 3y=ll (1) Adding (1) to (3): llx=77, or x=z7 Substituting this value for x in (1): 2-7— 3y=ll 14— 3y=ll — 3y= 11—14, or —3 y=l 3. Solve the following by the method of substitution: 9x + y=21 x + 9y=29 Answer yr=21— 9x Substituting this value for y in (2) : x + 9(21— 9x)=29 x+189— 81x=29 — 80x=:29— 189 — 80x=r— 160 x=2 Substituting this value for x in equation (2): x + 9y=29 2 + 9y=29 9y = 29— 2 9y=27 y=3 4. Solve the and subtraction: Answer following by addition 5x + 6y=r27 x + 7y=17 Answer Multiplying (2) by 5 : 5x+35y=85 (3) 5x + 6y=27 (1) Subtracting (1) from (3) : 29y=58 y=2 Substituting this value for y in equation (2): x + 7y=17 x+14=17 x=17— 14 x=3 5. Solve the following by the graphical method: 3x + 4y = 7 5x — 2y = 3 Answer Graphical solution appears below (Continued on next page) Graphical solution to Problem 5