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Transcript
Boardworks Maths Maths Algebra 2 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 9 © Boardworks Ltd 2014 Linear simultaneous equations An equation with two unknowns has an infinite number of solution pairs. For example: x+y=3 is true when x=1 and y=2 x = –4 and y=7 x = 6.4 and y = –3.4 We can represent the set of solutions graphically. y 3 The coordinates of every point on the line satisfy the equation. 0 2 of 9 and so on. x+y=3 3 x © Boardworks Ltd 2014 Linear simultaneous equations Similarly, an infinite number of solution pairs exist for the equation y–x=1 Again, we can represent the set of solutions graphically. There is one pair of values that satisfies both these equations simultaneously. This pair of values corresponds to the point where the lines x + y = 3 and y – x = 1 intersect: y y–x=1 1 -1 0 x x+y=3 This is the point (1, 2). At this point x = 1 and y = 2. 3 of 9 © Boardworks Ltd 2014 Linear simultaneous equations Two linear equations with two unknowns, such as x and y, can be solved simultaneously to give a single pair of solutions. When will a pair of linear simultaneous equations have no solutions? In the case where the lines corresponding to the equations are parallel, they will never intersect and so there are no solutions. Linear simultaneous equations can be solved algebraically using: The elimination method, or The substitution method. The solution to the equations can be illustrated graphically by finding the points where the two lines representing the equations intersect. 4 of 9 © Boardworks Ltd 2014 The elimination method If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example: Solve the simultaneous equations 3x + 7y = 22 and 3x + 4y = 10. Subtracting gives: 3x + 7y = 22 – 3x + 4y = 10 3y = 12 y=4 The terms in x have been eliminated. Substituting y = 4 into the first equation gives: 3x + 28 = 22 3x = –6 x = –2 5 of 9 © Boardworks Ltd 2014 The elimination method 6 of 9 © Boardworks Ltd 2014 The substitution method Two simultaneous equations can also be solved by substituting one equation into the other. For example: y = 2x – 3 2x + 3y = 23 Solve: 1 2 Call these equations 1 and 2 . Substitute 1 into 2 : 2x + 3(2x – 3) = 23 2x + 6x – 9 = 23 8x – 9 = 23 8x = 32 x=4 7 of 9 © Boardworks Ltd 2014 The substitution method Substituting x = 4 into 1 gives y=2×4–3 y=5 Check by substituting x = 4 and y = 5 into 2 : LHS = 2 × 4 + 3 × 5 = 8 + 15 = 23 = RHS So the solution is x = 4, y = 5. 8 of 9 © Boardworks Ltd 2014 Want to see more? This is only a sample of one of thousands of Boardworks Maths PowerPoints. To see more of what Boardworks can offer, why not order a full presentation, completely free? Head to: www.boardworks.co.uk/australiamathspresentation 9 of 9 © Boardworks Ltd 2014