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Published byMiguel Allen Modified over 2 years ago

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Open Sentences In Two Variables Objective: To find solutions of open sentences in two variables and to solve problems involving open sentences in two variables

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A solution of an open sentence with variables x and y is an ordered pair (2, -1) is a solution to the equation 3x + 2y = 4 because, 3(2) + 2(-1) = 4 The set of all solutions to the open sentence is called the solution set.

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Solve the equation y =4x – 6 if the domain of x is {-2, -1, 0, 1} x4x – 6ysolution -24(-2) – 6-14(-2,-14) 4(-1) – 6-10(-1,-10) 04(0) – 6-6(0,-6) 14(1) – 6-2(1,-2)

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You have one dollar in quarters and dimes. Find all the possible coin combinations. The problem asks for the number of quarters and dimes that make up one dollar. Let q equal the number of quarters and d equal the number of dimes The total value of the quarters and dimes will then equal:.25q +.10d

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.25q +.10d = 1.00 Solve for one variable -.25q +.25q +.10d = -.25q + 1.00.10d = -.25q + 1.00.10d/.10 = -.25q/.10 + 1.00/.10 d = -2.5q + 10 Keep two things in mind: both d and q must be whole numbers, q < 4 and d < 10

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Solution q-2.5q + 10d(q,d) 0-2.5(0) + 1010(0,10) 1-2.5(1) + 107.5ø 2-2.5(2) + 105(2,5) 3-2.5(3) + 102.5ø 4-2.5(4) + 100(4, 0)

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The solution set must then be: {(0,10), (2, 5), (4, 0)}

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