|
|
|||||
| Mathematics Magazine for Grades 1-12 |
|||||
|
|
|||||
|
|
|||||
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 = x2 + 2x 3 0 = x2 - 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 |
|||||
|
Read more on the written version of the publication. |
|||||