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Revision - Simultaneous Equations II Solving simultaneous equations using Substitution Method. By I Porter.

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Presentation on theme: "Revision - Simultaneous Equations II Solving simultaneous equations using Substitution Method. By I Porter."— Presentation transcript:

1 Revision - Simultaneous Equations II Solving simultaneous equations using Substitution Method. By I Porter.

2 Introduction. To solve simultaneous equations you can use a graphical method (not recommended), the algebraic methods of elimination or SUBSTITUTION. It is generally recommended to use one of the algebraic methods. Substitution Method. The substitution method is an algebraic technique that gives the exact solution of a pair of simultaneous equations. The following steps are generally required: Step 1:Use one of the equations to make one of the variables the subject. Step 2:Substitute the expression for the subject into the other equation. Step 3:Solve the new equation to find the value of the remaining variable. Step 4:Substitute this value into either of the original equations to find the other variable. Step 5:Write your answer, showing the value of both variables.

3 Example 1: Solve 3x + 2y = 8 and 5x - 3y = 7, by the substitution method. 3x + 2y = 8 (1) 5x - 3y = 7 (2) Select (1) or (2) to make either x or y the subject. Select (1) (why?) to make y the subject. From equation (1) This is one of the difficulties of this method, In most cases a fraction is formed. So, take extreme CARE! (3) Substitute equations (3) into equations (2). Expand the brackets. (4) Substitute equation (4) into (3) or (1) 3x + 2y = 8 (1) Hence, the solution is x = 2 and y = 1. Multiply by 2 for remove fraction.

4 Example 2: Solve 4x + 2y = 4 and 3x + 4y = 13, by the substitution method. 4x + 2y = 4 …(1) 3x + 4y = 13 …(2) Select (1) or (2) to make either x or y the subject. Select (1) (why?) to make y the subject. From equation (1) (3) Substitute equations (3) into equations (2). (4) Substitute equation (4) into (3) or (1) (3) Hence, the solution is x = -1 and y = 4. Expand the brackets. Simplify & solve for x.

5 Exercise: Solve the following simultaneous equation by the substitution method. 1) 5x - 2y = 18 & 3x - 4y = 8 2) 3x - y = 8 & 11x + 2y = 1 3) 5x + 2y = 3 & 2x - 3y = -14 4) 2x + 3y = -11 & 3x + 2y = -4 5) 5x + 8y = 40 & 2x + 3y = 30 Ans: x = 4, y = 1 Ans: x = 1, y = -5 Ans: x = -1, y = 4 Ans: x = 2, y = -5 Ans: x = 120, y = -70

6 Harder Example 1: Solve the following. When solving equations with different degrees (powers of x), the substitution method is the technique to use generally. 1) Solve the simultaneous equations y = x 2 + 4x + 3 & y = 5x + 9. In this case, use the linear equation to rearrange such that x or y is the subject. y = 5x + 9 … (1) has y as the subject already. Substitute (1) into the equation y = x 2 + 4x + 3 5x + 9 = x 2 + 4x + 3 Rearrange into a quadratic. x 2 - x - 6 = 0 Factorise or use quadratic formula to solve. (x - 3)(x + 2) = 0 x - 3 = 0 or x + 2 = 0 x = 3 (2) or x = -2 (3) Substitute the two values of x, (2) & (3) into equation (1). y = 5x + 9 y = 5(3) + 9y = 5(-2) + 9 y = 24y = -1 The solutions are x = 3 & y = 24 or x = -2 & y = -1

7 Harder Example 2: Solve the simultaneous equations y = x 2 - 5x + 10 & x - y = -2. Rearrange the linear equation, x - y = -2. y = x + 2 ….(1) Substitute equation (1) into the other equation. x 2 - 5x + 10 = x + 2 x 2 - 6x + 8 = 0 Rearrange into a quadratic. Factorise or use quadratic formula to solve. (x - 4) (x - 2) = 0 (x - 4) = 0 or (x - 2) = 0 x = 4 ….(2) or x = 2 …(3) Substitute the two values of x, (2) & (3) into equation (1). y = x + 2 ….(1) y = (4) + 2y = (2) + 2 y = 6y = 4 The solutions are x = 4 & y = 6 or x = 2 & y = 4

8 Harder Example 3: Solve 4x 2 + y 2 = 25 and x + y = 5 simultaneously. In this case, use x + y = 5. Make y the subject. y = 5 - x (1) Substitute (1) into the other equation. 4x 2 + y 2 = 25 4x 2 + (5 - x) 2 = 25 Expand and simplify. 4x x + x 2 = 25 5x x = 0 Factorise 5x(x - 2) = 0 Solve 5x = 0 or x - 2 = 0 x = 0 ….(2) or x = 2 …(3) Substitute the two values of x, (2) & (3) into equation (1). y = 5 - x (1) y = 5 - (0)y = 5 - (2) y = 5y = 3 The solutions are x = 0 & y = 5 or x = 2 & y = 3

9 Exercise: Solve by the substitution method. 1)y = 2x - 4 & y = x 2 - 2x - 9 2)y = x - 1 & xy = 6 3)y = 4x +12 & y = x 2 + 2x - 3 4)x - y = 9 & x 2 - y 2 = 27 5)y - 4x = 2 & y = 2x 2 + 3x - 8 Ans: x = 5, y = 6 or x = -1, y = -6 Ans: x = 3, y = 2 or x = -2, y = -3 Ans: x = 5, y = 32 or x = -3, y = 0 Ans: x = 6, y = -3 Ans: x = -2, y = -6 or x = -2.5, y = 12


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