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Only to be used for arranged hours Math 84 Activity # 9 Your Name: ___________________________ “Solving Linear Equations” An equation is equality of two algebraic expressions. Example. 5x + 2 5x + 2 = 3x – 10 . = 3x – 10 Subtract 3x from each side and record what you have on each side. Then subtract 2 from each side and record the result. Now divide both sides by 2. What is the result? We may think of an equation as a balance beam which we want to keep balanced all the time. It remains balanced if we take the same action on both sides. Therefore, given an equation A = B, we can perform the following operations on both sides. 1) A + C = B + C , for any real number C …………… Addition Property of Equality 2) A – C = B – C , for any real number C …………… Subtraction Property of Equality 3) A x C = B x C , for any real number C ………….. Multiplication Property of Equality ( but we don’t want to multiply by 0 – why? ) 4) A B , for any real number C ≠ 0 (why?) …… Division Property of Equality = C C Task 1:. A linear equation, in the variable x, is one that can be expressed as ax + b = c , where a, b, c are real numbers. Identify a, b, c in the linear equations below. 5x + 7 = 10 1 – x = 3 – 2x + 3 = 5 1 − 1 x = 0 4 –3 + 11x = 2 4 x = 5 3 2 5 x + = 1 3 4 0.5 x − 1.2 = 3.8 Only to be used for arranged hours A solution of an equation is a number that makes the equation a true statement upon substitution for the variable. Example. To determine whether – 2 is a solution of see if we get a TRUE statement. – 3x + 10 – 3(– 2) + 10 6 + 10 16 – 3x + 10 = 16, we substitute (– 2) for x and = 16 = 16 = 16 = 16 ……….TRUE Therefore – 2 is a solution (it is actually the solution) of the equation – 3x + 10 = 16 Try: Determine whether 3 is a solution of the equation 3x + (– 9) = 18 3.2 Using The Properties Of Equality To Solve Linear Equations. We will use the appropriate property of equality to isolate the variable. Examples: Solve the following equations: (Indicate the property used in each step). 1) x + 84 = 16 2) x – 84 = 16 x + 84 − 84 = 16 − 84 , subtraction − 68 x +0 = x = − 68 3) 1 x = 5 − 12 x − 84 + 84 = 16 + 84 , addition x = 100 4) 5 x = − 12 5x −12 = 5 5 multiplication , division 12 x = − 5 Task 2: In problems 5 through 8, which operation is used to solve for the variable? 3 1 9 5) 6) −3w = y = 2 4 2 2 3 2 1 −3w 9 1 ⋅ y =⋅ = ⋅ − 3 2 3 4 −3 2 3 3 1 ___________ ____________ w = − y= 2 6 1 5 ⋅ x = 5 ⋅ ( −12) , 5 x = − 60 7) x + ( −8) =11 x − 8 + 8 = 11 + 8 x = 19 _____________ 8) −15 + x = 24 15 + ( −15) + x= 24 + 15 x = 49 ___________________ Only to be used for arranged hours Observe that we have been performing the opposite operation to isolate the variable. Task 2: Now try the following. x 1 1) Determine whether (– 110) is a solution of = 220 2 2) Solve the following. Explain each step and check whether your solution is correct. a) 11 + x = – 22 b) x – 11 = – 22 c) 11x – 22 d) e) 11 x = 5 = − 22 x = 11 f) − − 22 5 x = 25 11 *In the following, you need to decide which property to use first. Check your answer and also discuss with team members. g) 1 − x =− 11 h) 1 − 2 x = −7 Only to be used for arranged hours Exercises: Solve the following equations. Justify your steps and check the solution. 1) −3x + 2 =− 22 2) 4 2 x − = 0 3 3 1 7 11 − x = − 2 3 6 4) 1 1 x − x = −x 3 2 3) 1 5) 2 x − + x = 3 7) 4 + x 3 7 2 11 1 x− x + = 12 5 5 5 5 9) 3( x – 5 ) + 2 = 4(x – 1 ) + 1 6) 0.3x + 1.5 ( x − 1) = 2.1 8) −2 x + 1 + 4 x =− 2 + 11x 2 10) 1 – x = x – 1