A brief note on Graphs and Consistency of Pairs of Equations . Read it before a test. Try each question at the end before you open its answer.
The graph of a pair A pair is a 1 x + b 1 y + c 1 = 0 a_1x + b_1y + c_1 = 0 a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 a_2x + b_2y + c_2 = 0 a 2 x + b 2 y + c 2 = 0 . Each equation draws a straight line. A solution of the pair is a point on both lines. Two points draw a line. Use a third as a check. Always check a solution in both equations. Three cases Lines cross: exactly one solution. The pair is consistent . Parallel lines: no solution. The pair is inconsistent . Coincident lines: infinitely many solutions. Consistent and dependent . A pair can never have exactly two solutions. The ratio test a 1 a 2 ≠ b 1 b 2 \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} a 2 a 1 = b 2 b 1 : one solution. a 1 a 2 = b 1 b 2 ≠ c 1 c 2 \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} a 2 a 1 = b 2 b 1 = c 2 c 1 : no solution. a 1 a 2 = b 1 b 2 = c 1 c 2 \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} a 2 a 1 = b 2 b 1 = c 2 c 1 : infinitely many solutions. Write both equations as a x + b y + c = 0 ax + by + c = 0 a x + b y + c = 0 first. To find k k k , set the ratios equal and check the third. A worked example For which k k k does 2 x + 3 y − 7 = 0 2x + 3y - 7 = 0 2 x + 3 y − 7 = 0 and 4 x + k y − 9 = 0 4x + ky - 9 = 0 4 x + k y − 9 = 0 have no solution?
Equal first two ratios. 2 4 = 3 k \dfrac{2}{4} = \dfrac{3}{k} 4 2 = k 3 , so k = 6 k = 6 k = 6 . Check the third ratio. − 7 − 9 = 7 9 \dfrac{-7}{-9} = \dfrac{7}{9} − 9 − 7 = 9 7 is not 1 2 \dfrac{1}{2} 2 1 . Decide. The lines are parallel, so k = 6 k = 6 k = 6 gives no solution. Practice with new numbers
A program to try The program decides the number of solutions without dividing, by cross-multiplying.
Run it. Do the three messages match the ratio test?
1. Change the last pair to (2, 3, -7, 4, 6, -13). What does the program print, and why?
Mistakes to avoid Saying "no solution" for coincident lines. They have infinitely many. Stopping at two ratios. Equal first two ratios mean parallel or coincident. Forgetting the standard form. The signs of the constants change the third ratio. Checking only one equation. A solution must fit both. Check yourself Do x + y = 6 x + y = 6 x + y = 6 and 2 x + 2 y = 12 2x + 2y = 12 2 x + 2 y = 12 have a solution? Stuck? Show a hint Show the answerClassify x − 2 y − 3 = 0 x - 2y - 3 = 0 x − 2 y − 3 = 0 and 2 x − 4 y − 5 = 0 2x - 4y - 5 = 0 2 x − 4 y − 5 = 0 . Stuck? Show a hint Show the answerA pair has parallel lines. What is the pair called? Stuck? Show a hint Show the answer