Algebra 8.6 Simple Equations

Transcription

1 Algebra 8.6 Simple Equations 1. Introduction Let s talk about the truth: 2 = 2 This is a true statement What else can we say about 2 that is true? Eample 1 2 = = 2 2 1= = = 2 4 = 4 You try - make a few statements about 4 that are true. Let s say you ve made a true statement: 2 = 2 What if something happens to one of your numbers? Add something as simple as a and your statement is no longer true. One way to respond is to accept what s happened to your number, and change your statement to make it true again: 2 + 1= This is called balancing: If you start with a true statement... a = b...and then something happens to one side... a + 1= b then, if you want the statement to stay true, you have to make the same thing happen to the other side. Algebra 8.6 Page 1

2 Introduction, Test yourself, continued Keeping a statement balanced and true is easy so long as you reflect any changes that happen to one side of the statement on the other side: Eample 2 = 2 + 1= = = 3(2) 2 = 2 2 y = 2y 2 = 2 2 = 2 If gets multiplied by 3, then 2 has to be multiplied by 3 as well... If is turned into a negative number, then 2 has to become a negative as well... a = 3 a + 2 = a 4 = 4a = a 4 = ab = a 3 = a = Make sure each statement stays true, no matter what happens to the a =3+2, 3 4, 4(3), 3 4, 4b, 33, 3 Here are some easy statements to keep balanced: Problem 1 If 3 = 3 then = Problem 2 If 4 = 4 then 4 9 = Problem 3 If = then + 2 = Problem 4 If = then 6 = Problem 5 If a = b then 3 a = Problem 6 If y = z then 4y 2 = But what if the statements are more complicated, for instance have more terms? + 6 = + 6 What if something happened to this statement? Algebra 8.6 Page 2

3 Factorising Balancing Test yourself, statements by grouping continued with (with more negative terms numbers), continued Eample 1 +2 If + 6 = = = + 8 If 2 gets added to one side, you have to add it to the other to keep the statement true. Once you ve done that, you can simplify by grouping the like terms. +3 If a + 4 = a + 4 a a+7=a+7 Eample 2 +5 If 2 = = = If a 3 = a 3 a a+3=a+3 Eample 3-6 If + 2 = = = 4-3 If a + 1= a + 1 a a 2=a 2 Eample 4 2 If + 3 = + 3 2( ) + 3 = 2( ) = If a + 2 = a + 2 3(a) + 2 Eample If + 3 = + 3 2( + 3) = 2( + 3) = If a + 2 = a + 2 3(a + 2) 3a+6=3a+6 3a+2=3a Eample If + 3 = = Since neither or 3 are divisible by 2, we can t simplify this statement. If a + 2 = a + 2 a a+2 3 = a Algebra 8.6 Page 3

4 Factorising Balancing Test yourself, equations by grouping continued(with negative numbers), continued So far you ve balanced statements where both sides are eactly the same: + 3 = + 3 But of greater interest are statements where both sides are not the same: + 3 = 5 If you made a statement like this, and something happened to change one side, how do you balance the other side to keep the statement true? Eample = = = 7 If 2 gets added to one side, you have to add it to the other to keep the statement true Once you ve done that, you can simplify by grouping the like terms +3 a + 4 = 6 a a+7=9 Eample = = = 1-7 a + 4 = 6 a a 3= 1 Eample y 2 = 4 2y = y + 2 = a 4 = 3 3a a+2=9 So far everything might seem easy. But consider this: Eample = 5 2( ) + 3 2(5) For instance: = 5 But 2(2) + 3 = 7 2(5) Here we are multiplying the by but doubling part of one side of the equation is not the same as doubling the whole of the other side. 3 a + 2 = 7 3(a) + 2 3(a)+2 3(7) Algebra 8.6 Page 4

5 Factorising Balancing Test yourself, equations, by grouping continued continued (with negative numbers), continued If you multiply or divide anything in an equation, you can only keep both sides balanced and true if you multiply and divide all terms on both sides the same way. Eample = 5 2( + 3) = 2(5) = 10 If you multiply one term by 2, then you have to multiply all terms by a + 3 = 7 2(a + 3) 2a+6=14 +4 Eample = = a + 7 = 12 a+7 5 =12 5 This is the hardest thing about balancing: making sure you do the same thing on both sides of an equation. Use the same kind of care you would use when defusing a bomb. Problem = 4 If you add 4 to one side then... Problem 2 +9 a + 3 = 7 a+12=16 +9=8 Problem = 18 Problem 4-6 y + 7 = 19 y +1=13 +8=14 Problem 5-7 m + 5 = 12 Problem 6-6 m 3 = 11 m 9=5 m 2=5 Algebra 8.6 Page 5

6 Factorising Balancing Test yourself, equations, by grouping continued continued (with negative numbers), continued Problem = 2 Problem 8 +6 a + 7 = 3 a+13=3 +8=2 Problem a = 5 Problem a = 3 8 a= 1 Problem = 12 Problem = 7 Remember: multiply all terms. Problem 13 Problem = 4 2m 4 = 11 6m 12= = = =5 Problem = 7 And divide all terms, too. Problem 16 6 y + 7 = 19 y +7 6 = = 7 4 Problem 17 6 m + 5 = 12 You ll be able to simplify this one. Problem 18 3 m 3 = 11 m 3 1=11 3 m+5 6 =2 11+a=2 Algebra 8.6 Page 6

7 Factorising Balancing Test yourself, equations, by grouping continued continued (with negative numbers), continued Problem = 6 Problem a = 4 2 a= =3 Problem a = 4 Problem m + 14 = 16 Problem = 6 2 Problem 24 3 y 3 4 = 6 y 12=18 +6=12 m+ 7 2 =4 1 a 2 = 2 3 Problem y 3 3 = 4 Problem 26 3y 2 3y 2 = 5 2 6y =15y 2y 9=12 Problem 27 a ab + 2 = 12 Problem 28 2m 2m 3 = 6 4m 2 6m=12m Problem 29 ^2 2 + = 3 This means raised to the power of 2 Problem 30 ^2 2 = = =9 a 2 b +2a=12a Algebra 8.6 Page 7

8 Factorising Balancing Test yourself, equations, by grouping continued continued (with negative numbers), continued Problem = 6 Problem = 13 Problem = 6 Problem = 2 Problem = 6 Problem = 4 Problem 37 n mn = 7 Problem 38 b ab = 5b Problem 38 2 = 9 Problem 39 3m 3m 3 + 3m = 15m Problem = 11 Problem = 15 =3 5 =15 m 2 +1=5 =9 a=5 m= 7 n =8 =2 =1 2 =2 =4 =2 Algebra 8.6 Page 8

9 Factorising Solving Test yourself, simple by grouping continued equations(with negative numbers), continued Why would you ever want to balance an equation? How about when you need to find the value of a variable: What number does this represent? + 2 = 4 You can solve this easily by deducting 2 from each side: = 4-2 = 2 Solving an equation means finding the hidden value of a variable The idea is to get the variable by itself on one side of the equation, and all the other information on the other side. This can mean using multiplication and division too: Eample 1 2 = = 8 = 4 2 2a = 6 a=3 Sometimes you can solve an equation in one step: Eample 2 3 = 4 3 = 4 = a 2 = 5 a=10 Often it will take more than one step: Eample = = = 12 = 4 2a + 5 = 17 a= Algebra 8.6 Page 9

10 Solving Test yourself, simple continued equations of the form +1=2 Eample 1 5 = = 8 = a 5 = 7 a=12 Eample 2 6 = = 4 = 2 +6 a 6 = 2 a=4 Problem = 7 Problem 2 m + 12 = 10 Problem = 9 Problem 4 6 = 13 Problem = 0 Problem 6 5 = 2 Problem 7 3 = 3 Problem 8 3 = 3 Problem = 3 Problem = 15 =27 =0 =0 =6 =7 = 5 =19 =4 m= 2 =4 Algebra 8.6 Page 10

11 Factorising Substituting Test yourself, by numbers grouping continued to check (with negative your answers numbers), continued Seriously, when doing this kind of algebra there is no ecuse for getting it wrong because you can always check your work by substituting numbers for variables: Eample 1 3 = 3 = = 3 3 = 3 Plug the answer back into the original and check you get the same result. +5 a 5 = 7 +5 a = 12 That was an easy eample. Here s what s coming: Eample = = = 10 = 2 5( 2) 3 = = = After all that, you ll really want to check that you didn t muck up anything. That s good. 5 4 = 14 Or try this one on for size: Eample = = = 4 = = = 6 6 = Plugging back in = 6 Always check your work by plugging your answer back into the question. Algebra 8.6 Page 11

12 Negative Factorising Test yourself, numbers by grouping continued(with negative numbers), continued What do you do in this situation: 3 = 6 = 3 Do you leave the answer as -? Usually, and this is only a rule of thumb, when solving an equation you want to find the value for the positive version of the variable, not the negative. 3 = 6 = 3 = 3 This is better How do you get from one to the other? One way is to just remember that you can flip positive and negative numbers around like this: Eample 1 y = 5 y = 5 a = 12 a= 12 Eample 2 y = 5 y = 5 a = 4 a=4 But if you want to know why you can do this, it s simple: you re just multiplying (or dividing) both sides of the statement by -1: Eample 3 3 = 6 = 3 1( ) = 1(3) = 3 If you understand this step, it will make your life with negative numbers much easier. 7 a = 9 a= 2 Eample 4 3 = 6 = 9 1( ) = 1( 9) = 9 7 a = 9 a=16 Algebra 8.6 Page 12

13 Solving Test yourself, simple continued equations of the form 2=4 Eample 1 3 = = 12 = 4 3 4a = 12 a=3 Eample 2 5m = 20 (-5) 5m = 20 m = 4 Problem 1 4 = 24 (-5) 5m = 30 Problem 2 3 = 21 m= 6 Problem 3 7 = 35 Problem 4 6 = 12 Problem 5 3 = 8 Problem 6 = 100 Problem 7 8 = 24 Problem 8 m 2 = 7 Problem 9 2 = 5 Problem 10 6 = 2 =12 =10 m= 14 =3 =100 = 24 =2 =5 = 7 =6 Algebra 8.6 Page 13

14 Solving Test yourself, simple continued equations of the form 2+1=5 Eample = = = 10 = a 3 = 13 a= 5 Eample = = 9 (-3) 3 = 3-6 = 1 (-3) 5m = 30 Problem = 17 Problem = 16 Problem y = 17 Problem 4 6 3a = 18 Problem = 9 Problem 6 6y 7 = 19 Problem = 33 Problem = 1 =3 = 3 y = 2 = 1 a= 4 y =6 =3 =5 a= 6 Algebra 8.6 Page 14

15 Solving Test yourself, simple continued equations of the form 2+1=5, continued Problem 9 6 4m = 10 Problem = 2 Problem = 13 Problem = 7 Problem a = 17 Problem = 12 =3 a=5 = 4 =2 =2 m=1 Algebra 8.6 Page 15

16 Test Solving yourself, simple continued equations of the form 2+1=3 Eample = = = 2 = Always take care with negative numbers! a 2 5 = 3 a=4 Eample = = = 4 = y = 6 Problem = 8 Problem 2 a 2 8 = 5 Problem 3 y 3 8 = 5 Problem 4 m = 13 m=14 y =9 a=6 =16 y =16 Problem = 3 Problem = 7 = 3 Algebra 8.6 Page 16 y = 15

17 Solving Test yourself, simple continued equations of the form 2+1=+2 What if you have a variable on both sides of the equation? Eample = = = = 8 = Removing the 5 from the left side......and removing 2 from the right side 4a + 1= a + 13 a=4 Remember, the point is to get one variable by itself on one side of the equation. Eample = = = = 8 = a 14 = 4a 2 a=2 Problem = + 10 Problem = Problem = 3 32 Problem = =9 = 14 =3 =2 Algebra 8.6 Page 17

18 Solving Test yourself, simple continued equations of the form 2+1=+2, continued Problem = 5 5 Problem = Problem 7 a 4 = 3a 8 Problem 8 y + 6 = 6y 9 Problem 9 3 = 4 9 Problem 10 8 = Problem = 11 4 Problem = 7 12 =5 = 1 = 3 =2 y =3 a=2 = 1 = 3 Problem = 5 10 Problem = =0 = 10 Algebra 8.6 Page 18

19 12. Test Solving yourself, equations continued of the form 2(+1)=4 Eample 1 2( + 3) = = 10 2 = 4 = 2 2(y + 4) = 12 y =2 Eample 2 3(2 4) = = 18 6 = 30 = 5 4( 5) = 28 Problem 1 2(3 4) = 10 Problem 2 2( + 6) = 6 Problem 3 5(2 + 9) = 15 Problem 4 2(5 6) = 22 Problem 5 3( + 4) = 6 Problem 6 3(2 1) = 9 =2 = 2 = 1 = 3 = 3 =3 =12 Algebra 8.6 Page 19

20 12. Test Solving yourself, equations continued of the form 2(+2)=3(+1) Eample 1 4( + 3) + 3( 2) = = = 34 7 = 28 = 4 Group like terms... 3(2 3) + 2( 4) = 25 Eample 2 2( 7) = 6( + 1) 2 14 = = = 20 = 5 3( + 4) = 2(4 + 1) =2 = 1 Problem 1 2(3 4) + 3( + 4) = 41 Problem 2 5(2 + 1) 3( 3) = 35 Problem 3 3(3 4) 2(2 3) = 11 Problem 4 3(3 + 4) = 2( + 13) Problem 5 7( + 2) = Problem 6 4( + 3) = 2( + 6) = 4 = 1 =2 = 1 =3 = 5 Algebra 8.6 Page 20

21 Test Mied yourself, practice continued Problem = 2 Problem = 12 Problem = 3 Problem 4 5 = 8 = 3 =5 = 7 Problem 5 2 = 6 Problem 6 3 = 4 =12 = 12 = 8 Problem 7 8 = 48 Problem 8 = 1 =6 Problem = 7 Problem 10 a = 8 Problem 11 m 2 5 = 8 Problem = 5 =20 m= 6 a= 3 = 8 =1 Algebra 8.6 Page 21

22 Test Mied yourself, practice, continued Problem 13 2(2 + 3) = 20 Problem 14 11( 4) = 33 Problem 15 3( + 3) = 27 Problem 16 3(y 6) = 4(y + 4) = 12 =7 Problem 17 5(2 + 3) 4( + 2) = 19 Problem 18 2(4 3) = 3( 7) = 3 =2 y = 34 = Algebra 8.6 Page 22