E x a mp l e 1 10 Sketch the graphs of the following equations 3 2 2 6 x y x y

E x a mp l e 1 10 sketch the graphs of the following

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E x a mp l e 1 . 10 Sketch the graphs of the following equations. 3 2 2 6 x y x y   S o l u t i o n As shown in the opposite graph . Ac Activity x.x Spend at least 20 -30 minutes on this activity. State whether the following system of linear equations are coincident, parallel or intersect. Explain your answers. a) 2 5 4 2 10 x y x y   b) 2 3 6 4 6 0 x y x y c) 2 5 4 x y x y 3 3 x y Activity 1.3
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Unit 1: Basic Algebra and Linear Equations 1.15 1.4 Solving systems of linear equations There are many techniques of solving systems of linear equations. These include graphical methods, algebraic methods and elementary row operation. However in this course, we will be dealing with algebraic methods only. Algebraic methods are where algebraic operations are performed on linear equations to yield an equation with only one variable and then solve the new equation. The rules for obtaining any two equivalent equations are shown below. Equation Operations Producing Equivalent Systems Equivalent systems of equations are systems that have the same solution set . There are two algebraic methods to study and these are Substitution and Elimination Methods . Substitution Method E x a mp l e 1 . 11 2 3 3 (1) Solve 4 (2) x y x y   Summary of Steps 1. Make one variable the subject, for one of the equations. 2. Substitute the resultant equation in the remaining equations. 3. If we obtain an equation with only one variable, solve it. Otherwise, repeat steps 1 and 2 until such an equation is obtained. 4. Find the values of the remaining variables by back-substitution. 5. Check the solution. Theorem: A system of linear equations is transformed into an equivalent system if: (1) Two equations are interchanged. (2) An equation is multiplied by a nonzero constant. (3) A constant multiple of one equation is added to another equation.
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Unit 1: Basic Algebra and Linear Equations 1.16 S o l u t i o n Step 1: Making the subject of equation (2) we have 4. y y x Step 2: We substitute this result in equation (1) and simplify, we have 2 3( 4) 3 2 3 12 3 x x x x Step 3: Since now we have only one variable ( ), we solve for 5 15 3. x x x x Step 4: Knowing 3, we can find by substituting the value of into either (1) or (2). Note that both give the same answers. x y x (This technique is also known as back substitution.) In here we are using equation (2). 4 4 3 4 1. x y y x       2(3) 3( 1) 3 Step 5: Check ( 1) 3 4     E x a mp l e 1 . 12 Solve the following system by substitution method. 36 9 (1) 3 12 (2) 3 y x y x S o l u t i o n Substitute (1) into (2), we have 1 3 (36 9 ) 12 3 3 12 3 12 12 12 x x x x This result does not give a value of either y nor x . However it gives a true statement.
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