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Solve for x by dividing both sides by 7: 7 x 21 →x=3 = 7 7 4. Substituting back into either of the original equations, solve for y: 2x + y = 8 → 2(3) + y = 8 → y = 8 – 2(3) = 8 – 6 → y = 2 5x – y = 13 → 5(3) – y = 13 → –y = 13 – 15 = –2 → y = 2 5. solution: (3, 2) 6. 22b Equations, Inequalities & Linear Programming Solve the given system of linear equations through algebraic elimination: x + 3y = 5 x – 5y = –3 1. The easiest variable to eliminate is the x. 2. Subtracting straight down: 8y = 8 3.
Although there will be two equations, the finalized graph will look as if there was only one line drawn. ) 7KHVHĆOLQHVĆLQWHUVHFWĆZLWKĆHDFKĆRWKHUĆ 7KHVHĆOLQHVĆDUHĆSDUDOOHOĆ 7KHVHĆOLQHVĆFRLQFLGHĆ GLDJUDP GLDJUDP GLDJUDP Here are the equations that produced the graphs above: 5x + 4y = 20 4x – 5y = 20 6x – 4y = 12 3x – 2y = 18 5x – 2y = 10 10x – 4y = 20 slope = –5/4 (0, y) = (0, 5) (x, 0) = (4, 0) slope = 4/5 (0, y) = (0, –4) (x, 0) = (5, 0) slope = 6/4 (0, y) = (0, 3) (x, 0) = (2, 0) slope = 3/2 (0, y) = (0, –9) (x, 0) = (6, 0) slope = 5/2 (0, y) = (0, –5) (x, 0) = (2, 0) slope = 10/4 (0, y) = (0, –5) (x, 0) = (2, 0) We can analyze what elements of the linear equations will produce each of the possible interactions.
5. Use the fractional slope to determine the needed vertical (rise) and horizontal (run) moves off of the y-intercept. For this example, we will need to move 4 units down and 3 units to the right from the y-intercept of (0, 4). Make a mark at this position: (0 + 3, 4 – 4) or E (3, 0). 6. Connect the y-intercept point with the point that was determined after the “rise over run” movements (in step #5) with a nice long straight line that extends to the outer portion of the coordinate grid.