Mathematics Magazine for Grades 112 





Theory:Carnots Theorem
Solutions from the Previous Issue:
1. What is the maximum number for the possible points of intersection of a circle and a triangle? Solution: There are 6 possible points of intersection of a circle and a triangle. 2. Solve for x: Solution: The procedure is to multiply both sides of the equation by the LCD, which in this case is (x 2)(x + 1). It is essential to multiply every term on both sides of the equation. In so doing, all denominator cancel.
x + 1 2(x 2) = 7 x + 1 2x + 4 = 7 x = 2 x = 2 3. Solve for x: Solution:
The LCD is (x 3)((x 2). Multiplying both sides by this yields 3(x  2) 2(x 3) = 3 3x 6 2x + 6 = 3 x = 3 4. Solve for x: Solution: We multiply both sides by the LCD, (x 1)(x + 3), to obtain x + 3 + 3(x 1) = (x 1)(x + 3) x + 3 + 3x 3 = x^{2} + 2x 3 0 = x^{2}  2x 3 (x 3)(x + 1) =0 x = 3; x = 1 Since the LCD is nonzero for each of these, the solution set is {1,3} Proposed Exercises
Solve each system of equations 1.
16x  y = 283 2.
2x + 9y = 219 3.
8x + 13y = 44 4.
15x + 17y = 327 

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