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Simultaneous equations Applying algebraic skills to linear equations I can… …solve simultaneous equations graphically …solve simultaneous equations algebraically.

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Presentation on theme: "Simultaneous equations Applying algebraic skills to linear equations I can… …solve simultaneous equations graphically …solve simultaneous equations algebraically."— Presentation transcript:

1 Simultaneous equations Applying algebraic skills to linear equations I can… …solve simultaneous equations graphically …solve simultaneous equations algebraically by substitution …solve simultaneous equations algebraically by elimination …use context to create simultaneous equations

2 An introduction Reminders The equation can be represented by a straight line which has a of m and passes through the point A System of Equations consists of two (or more) equations with at least two variables. These are also referred to as as their solution holds true for both equations. Systems of Equations * by drawing graphs* by substitution* by elimination When the System consist of two equations, with two variables, there are three methods of finding the solution: - y = mx + c gradient(0, c) Simultaneous Equations

3 …solve simultaneous equations graphically If the lines representing the equations are drawn then the solution is the coordinates of the point where the lines intersect (meet). Either: Example Solve these equations simultaneously y = ½x + 1 & y = 7 – x Line one - y = ½x + 1 When x = 0, y = → (0, 1) When x = 2, y = → (2, 2) Line two - y = 7 – x intercept = m = Lines intersect at so solution is x = y = Drawing straight lines - set x = 0, find the y-coordinate from the formula, then set y = 0 and find x - pick 2 values for x and find the corresponding values of y - use the y-intercept, gradient and y = mx + c Draw the two lines with the given information ½ × 0 + 1 = 1 ½ × 2 + 1 = 2 (0, 7) (4, 3) 4 3

4 …solve simultaneous equations graphically (continued) Simultaneous equations are often used to solve problems and use letters other than x & y. Solving simultaneous equations graphically will often only give approximate solutions and relies on accurate drawing of graphs. In order to get precise solutions it is better to use one of the other methods – substitution or elimination.

5 …solve simultaneous equations algebraically by substitution At the point where the lines meet, the values of x and y are in both equations. This allows the first equation to be substituted into the second. Examples 1) y = x + 1 y = 4x – 5 Replace the y in the second equation with the first = 4x – 5 Substitute x into the first equation to find y x = 2, y = the solution is Check by substituting into the second equation, if it is true the solution is correct the same Make x the subject of the formula 3 = 4 × 2 – 5 x + 1 1 = 3x – 5 6 = 3x x = 2 2 + 1 = 3 2) y = 4x + 1 2y – 5x + 4 = 0 2 – 5x + 4 = 0 (4x + 1) 8x + 2 – 5x + 4 = 0 3x + 6 = 0 3x = –6 x = –2 x = -2, y = 4 × (–2) + 1 = – 7 the solution is (2, 3) (–2, –7) 2 x (-7) - 5 x (-2) + 4 = 0

6 …solve simultaneous equations algebraically by substitution Sometimes it is necessary to rearrange one of the equations first. 3) y – 2x = 3 3y – 2x = 17 Rearrange the first equation y = 6x + 9 – 2x = 17 4x + 9 = 17 4x = 8 x = 2 x = 2, the solution is Example 2x + 3 Replace the y in the second equation with the first Substitute x into the first equation to find y Check by substituting into the second equation, if it is true the solution is correct Make x the subject of the formula 3 – 2x = 17(2x + 3) y – 2 × 2 = 3 y – 4 = 3 y = 7 (2, 7) 3 x 7 – 2 x 2 = 17

7 …solve simultaneous equations algebraically by elimination In this method the equations are added or subtracted so that one of the variables will be eliminated. 1) 2x + 3y = 35 7x – 3y = 1 Place the letters in the same order Substitute into the first equation to find y 2 + 3y = 35 the solution is 2) 3x + 2y = 7 5x + 2y = 13 3 + 2y = 7 the solution is Examples Add/subtract to remove a letter Solve for the remaining letter 2x + 3y + 7x + (-3y) = 9x 35 + 1 = 36 So 9x = 36 x = 4 x 4 3y = 27 y = 9 (4, 9) 3x + 2y – 5x – 2y = –2x 7 - 13 = – 6 So -2x = -6 x = 3 × 3 2y = –2 y = –1 (3, –1)

8 …solve simultaneous equations algebraically by elimination Sometimes it is necessary to multiply one or both of the equations before adding or subtracting. 3) 3x + 4y = 26 6x – y = 7 4(6x – y) = 4 x 7 the solution is 4) 6x + 2y = 38 2x – 3y = 20 the solution is Place the letters in the same order Substitute into the first equation to find y Examples Add/subtract to remove a letter Solve for the remaining letter Multiply as required 3x + 4y + 24x + (-4y) = 27x 26 + 28 = 54 So 27x = 54 x= 2 3 + 4y = 26 4y = 20 y = 5 (2, 5) x 2 24x – 4y = 28 3(2x – 3y) = 3 x 20 6x – 9y = 60 6x + 2y - 6x – (-9y) = 2y + 9y = 11y 38 – 60 = -22 So 11y = -22 6x + 2 = 38 y = -2 × (–2) 6x = 42 x = 7 (7, –2)

9 …solve simultaneous equations algebraically by elimination 5) 3x + 2y = 7 4x + 3y = 9 9x + 6y - 8x – 6y = x 21 – 18 = 3 So x = 3 3 + 2y = 7 the solution is Place the letters in the same order Substitute into the first equation to find y Example Add/subtract to remove a letter Multiply as required3(3x + 2y) = 3 x 7 2(4x + 3y) = 2 x 9 9x + 6y = 218x + 6y = 18 × 3 2y = –2 y = –1 (3, –1)

10 …use context to create simultaneous equations Some problems can be solved using simultaneous equations by turning the problem into a set of equations. If answering a question set in a particular context you must write your final answer in context. 1) A jug and two glasses hold 1·6 litres altogether. Two jugs and three glasses hold 2·9 litres altogether. How much does each hold? j + 2g = 1·6 2j + 3g = 2·9 2j + 4g – 2j – 3g = g 3.2 – 2.9 = 0·3 So g = 0.3 j + 2 × = 1·6 So a Jug holds 1 litre and a glass holds 0·3 litres Examples Choose appropriate letters Substitute into the first equation to find y Add/subtract to remove a letter Multiply as required 2j + 4g = 3·2 2(j + 2g) = 2 x 1.6 0·3 j + 0.6 = 1.6 j = 1

11 Simultaneous equations Applying algebraic skills to linear equations I can… …solve simultaneous equations graphically …solve simultaneous equations algebraically by substitution …solve simultaneous equations algebraically by elimination …use context to create simultaneous equations

12 An introduction Reminders The equation can be represented by a straight line which has a of m and passes through the point A System of Equations consists of two (or more) equations with at least two variables. These are also referred to as as their solution holds true for both equations. Systems of Equations * by drawing graphs* by substitution* by elimination When the System consist of two equations, with two variables, there are three methods of finding the solution: -

13 …solve simultaneous equations graphically If the lines representing the equations are drawn then the solution is the coordinates of the point where the lines intersect (meet). Either: Example Solve these equations simultaneously y = ½x + 1 & y = 7 – x Line one - y = ½x + 1 When x = 0, y = → (0, 1) When x = 2, y = → (2, 2) Line two - y = 7 – x intercept = m = Lines intersect at so solution is x = y = Drawing straight lines - set x = 0, find the y-coordinate from the formula, then set y = 0 and find x - pick 2 values for x and find the corresponding values of y - use the y-intercept, gradient and y = mx + c Draw the two lines with the given information

14 …solve simultaneous equations graphically (continued) Simultaneous equations are often used to solve problems and use letters other than x & y. Solving simultaneous equations graphically will often only give approximate solutions and relies on accurate drawing of graphs. In order to get precise solutions it is better to use one of the other methods – substitution or elimination.

15 …solve simultaneous equations algebraically by substitution At the point where the lines meet, the values of x and y are in both equations. This allows the first equation to be substituted into the second. Examples 1) y = x + 1 y = 4x – 5 Replace the y in the second equation with the first = 4x – 5 Substitute x into the first equation to find y x = 2, y = the solution is Check by substituting into the second equation, if it is true the solution is correct Make x the subject of the formula 2) y = 4x + 1 2y – 5x + 4 = 0 2 – 5x + 4 = 0 x = -2, y = the solution is

16 …solve simultaneous equations algebraically by substitution Sometimes it is necessary to rearrange one of the equations first. 3) y – 2x = 3 3y – 2x = 17 Rearrange the first equation y = x = 2, the solution is Example Replace the y in the second equation with the first Substitute x into the first equation to find y Check by substituting into the second equation, if it is true the solution is correct Make x the subject of the formula 3 – 2x = 17

17 …solve simultaneous equations algebraically by elimination In this method the equations are added or subtracted so that one of the variables will be eliminated. 1) 2x + 3y = 35 7x – 3y = 1 Place the letters in the same order Substitute into the first equation to find y 2 + 3y = 35 the solution is 2) 3x + 2y = 7 5x + 2y = 13 3 + 2y = 7 the solution is Examples Add/subtract to remove a letter Solve for the remaining letter

18 …solve simultaneous equations algebraically by elimination Sometimes it is necessary to multiply one or both of the equations before adding or subtracting. 3) 3x + 4y = 26 6x – y = 7 4(6x – y) = 4 x 7 the solution is 4) 6x + 2y = 38 2x – 3y = 20 the solution is Place the letters in the same order Substitute into the first equation to find y Examples Add/subtract to remove a letter Solve for the remaining letter Multiply as required 3 + 4y = 26 3(2x – 3y) = 3 x 20 6x + 2 = 38

19 …solve simultaneous equations algebraically by elimination 5) 3x + 2y = 7 4x + 3y = 9 3 + 2y = 7 the solution is Place the letters in the same order Substitute into the first equation to find y Example Add/subtract to remove a letter Multiply as required3(3x + 2y) = 3 x 7 2(4x + 3y) = 2 x 9

20 …use context to create simultaneous equations Some problems can be solved using simultaneous equations by turning the problem into a set of equations. If answering a question set in a particular context you must write your final answer in context. 1) A jug and two glasses hold 1·6 litres altogether. Two jugs and three glasses hold 2·9 litres altogether. How much does each hold? j + 2 × = 1·6 So a Jug holds 1 litre and a glass holds 0·3 litres Examples Choose appropriate letters Substitute into the first equation to find y Add/subtract to remove a letter Multiply as required 2(j + 2g) = 2 x 1.6


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