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1. Math 084 - Quiz 4 Solutions (1) Below are two lines that form a linear system. (a) List all solutions of the system. A solution to a linear system looks like an intersection point of the lines. Looking at the graph, the intersection point is (−2, 3). (b) Write the system of linear equations that correspond to this system. The first line contains the points (−2, 3) and (4, 4). So the slope of the line is 4−3 = 16 . 4 − (−2) Now we can use this slope and the point (4, 4) to write the equation of the first line in point-slope form: y − 4 = 16 (x − 4) y = 16 x + ⇒ 10 3 . The second line contains the points (−2, 3) and (3, 1). So the slope is 1−3 = − 25 . 3 − (−2) Now we can use this and the point (3, 1) to write the equation: y − 1 = − 52 (x − 3) y = − 25 x + ⇒ 11 5 . So the system of equations is y = 16 x + 10 3 y = − 25 x + 11 5 . (MORE ON BACK) 1 2 (2) Solve(the linear systems below. x + 2y = 8 (a) x − 3y = 3 Since both equations in this system are in standard form, I’ll use elimination. First, I’ll multiply the second equation by −1, which gives me ( x + 2y =8 −x + 3y = −3 Now the x’s are set up to cancel, so I’ll add the equations to get 5y = 5 ⇒ y = 1. Now I’ll put 1 in for y in the equation x + 2y = 8: x + 2(1) = 8 ⇒ x = 6. So the solution to the system of equations is (6, 1). ( y = 4x + 5 (b) −8x + 2y = 2 Since the first equation is already solved for y, I’ll use substitution. Putting 4x + 5 into the second equation for y, I get −8x + 2(4x + 5) = 2 −8x + 8x + 10 = 2 10 = 2. There is no value of x that makes this last line true, so there are no solutions to the system of equations.