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Solving simple problems by forming and solving equations

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MATHEMATICS

Grade 9

NUMBER PATTERNS, GRAPHS, EQUATIONS,

STATISTICS AND PROBABILITY

Module 16

SOLVING SIMPLE PROBLEMS BY FORMING AND SOLVING EQUATIONS

ACTIVITY 1

To solve simple problems by forming and solving equations

[LO 2.2, 2.4]

How do we solve this problem?

Jamie’s dad is four times as old as Jamie. His father is 40. How old is Jamie?

a) We could think really hard, and try a few guesses. For instance: If Jamie is 1 year old, then his father must be 4. Not correct. What about 2 years old? And so on.

b) Make a table: Complete the empty spaces. Does this help?

Table 1
Age 1 2 3 4    
Age x 4 4 8 12   28 40

c) Draw a flow diagram: Is this useful?

Figure 1
Figure 1 (Picture 1.png)

  • As you can see, there is not much difference between the three methods – you can use any one that suits you. But they are not really very useful, because you have to do a lot of guessing. If the problems become very difficult, these methods become impossible to use.
  • So we will use a better way.
  • The way to do this problem is to make an equation from the information, and then to solve the equation and use the solution to write down an answer.
  • Making equations is not very easy to start with, but it gets a lot easier with practice.
  • An equation needs an equal sign as well as a variable (we usually use x, but it can be any letter).

Jamie’s dad is four times as old as Jamie. If his father is 40, how old is Jamie?

  • This is the general form of the equation:
  • Four times Jamie’s age = father’s age, which is 40.
  • The question asks Jamie’s age; so we let Jamie’s age be x.
  • Now the equation becomes: 4x = 40
  • and this asks us to find a number which gives 40 when multiplied by 4. If we divide 40 by 4, we get 10. So x must be 10, and therefore Jamie is ten years old. It is easy to see that this answer is correct.

A problem for you to do: Leon has 48 marbles. Amy has only a third as many as Leon; how many does she have? Use the last method.

The way you should set out your answer is:

State what the variable represents.

  • Write an equation using the variable.
  • Solve the equation.
  • Write down the solution, namely what the value of the variable is.
  • Write down the answer to the problem in words.

Please note that the first and last steps are in ordinary words, and the middle step(s) in algebra.

Exercise:

Find the answers to the following problems:

1. Mr Jacobs has R295,45 in his pocket. Mrs Jacobs has R55,30 less than her husband in her purse. How much money does she have in her purse?

2. I think of a number. I multiply it by 7 and divide the answer by three. I get 49. What was the number I first thought of? Remember to check your answer.

3. In America Joanie buys an item that is marked $5,75. She works out that it would be R41,69 when she converts the dollars to rand. What is the rand/dollar exchange rate she used?

ACTIVITY 2

To develop effective methods for solving more complicated equations

[LO 2.3, 2.4]

  • We will be doing more word problems, but we have to concentrate more on the algebra in our solutions. Make sure that you know exactly what each of the steps means and why it is done.
  • Some of the following problems are followed by the solutions. But try to find the answer without looking ahead. Then compare your answer and the way you set it out with the answer.

1. If I treble the price of the CD I bought, and then add R90 to the answer, I get the price of the R495 portable CD player I bought at the same time. How much did the CD cost? How much did I spend at the time?

  • Solution:Cost of CD × 3 plus R90 = price of CD player

Let x be the price of the CD.

x × 3 + 90 = 495 Translate the words into algebra

3x + 90 = 495 simplify

3x = 495 – 90 Do the same (– 90) on both sides of the equal sign

3x = 405 simplify

x = 405  3 Do the same ( 3) on both sides of the equal sign

x = 135 simplify

The CD cost R135.

I spent R135 + R495 = R630 altogether.

2. Mrs Williams is a supervisor in a factory. Devon Jones has just been appointed as a trainee supervisor at a weekly wage of R900. She has heard that if R535 is subtracted from her weekly wage, then the remaining amount is exactly half of the new man’s weekly wage. Mrs Williams is concerned that she is not earning as much as the less experienced young man. Is she right to worry? Hint: Let y be Mrs Williams’s wage.

3. There is a number which is 6 bigger than another number. The sum of the two numbers is 28. What is the number?

  • Solution:

(A number) + (the number – 6) = 28

Let the number be x; then the other number is x– 6

x + (x–6) = 28 Translate into algebra

x+ x – 6 = 28 Remove brackets

2x – 6 = 28 simplify

2x = 28 + 6 Add 6 to each side

2x = 34 simplify

x = 34  2 Divide each term by 2

x = 17 simplify

The number is 17

Check the answer!

4. Alan and his sister walk straight to school every morning. In the afternoon Alan walks straight home again, but his sister always walks home past her friend’s house; this route is twice as long as the route Alan takes home. Together their morning and afternoon distances add up to 1½ kilometre. How far is their house from the school?

It is important to be able to translate the problem into an algebraic equation, and it is also important to be able to do the algebra necessary to solve the equation, and so to solve the problem. We will give that some attention now. Here are the steps in the solution to problem 3 given above.

  • First we remove brackets and simplify: x + (x–6) = 28 x + x – 6 = 28
  • Then we make sure all the terms with the variable are on the left of the equal sign, and all the terms without are on the right of the equal sign, by adding or subtracting terms on both sides as necessary, and simplifying: 2x – 6 = 28 2x = 28 + 6 2x = 34
  • If the variable has a numerical coefficient, we divide both sides by it, and simplify: x = 34  2 x = 17

We are using all the skills we learned when we simplify e x pressions.

Practise these skills on the following algebraic equations:

5. (a) 5x = 35 (b) 4x = 22 (c) 3x – 90 = 0 (d) ½ x = 21

6 (a) 5x + 15 = 35 (b) 8 + 4x = 22 (c) 3x – 90 = –60 (d) ½ x + 3 = 15

7 (a) 5x + 15 = 2x (b) 8 + 4x = 22 – 2x (c) 3x – 90 = x (d) ½ x + 3 = 4 – ½ x

8 (a) 5(x + 1) = 20 (b) 8 + 4(x – 1) = 0 (c) x(x + 3) = x2 + 6 (d) ½ (4x + 6) = 1

9 (a) 2(x + 1) = x + 2 (b) 2(x + 3) = 2x + 6 (c) 3 – 2x = –2(1 + x)

Now we must make sense of the solutions we found in these equations.

Here are some answers:

8 (a) x = 3 (b) x = –1 (c) x = 2 (d) x = –1

  • These are acceptable answers.
  • They give the value of the variable (x), which makes the equation true.

9. (a) x = 0 This is also an acceptable answer.

(b) 2x + 6 = 2x + 6 2x – 2x = 6 – 6 0 = 0

  • This answer does not give us a single value of x.
  • But the statement is true: zero is equal to zero.
  • When we get an answer stating an obvious truth, like 12 = 12 or –3 = –3, etc., then we know that any value of the variable will make the equation true.
  • So the answer to give: x can take any value.

(c) 3 – 2x = –2 – 2x –2x + 2x = –2 – 3 0 = –5

  • This answer does not give a value for x.
  • The statement is in fact untrue. Zero is not equal to negative five.
  • When we get an answer which is untrue, like 5 = –5 or 2 = –9, etc., then we know that no value of the variable will make the equation true.
  • So we give the answer: There is no solution.

From now on, look out for these special cases (you won’t see them often) and give the appropriate answer.

ACTIVITY 3

To confirm that solutions are correct

[LO 2.4, 2.6]

  • For many problems in mathematics it is very difficult to know whether our answers are correct, but when we solve equations it is very easy. We simply check our answer! This has to be done very carefully, in a specific form.

This is how: Let’s go back to question 8 above.

(a) 5(x + 1) = 20 gives a solution: x = 3

We start with the original equation.

Check the left hand side (LHS) and right hand side (RHS) separately.

Substitute the solution for x and simplify:

LHS = 5(x + 1) = 5[(3) + 1] = 5(3 + 1) = 5(4) = 20

As usual, using brackets when substituting is very helpful.

RHS = 20

Because the RHS and LHS are equal, we know the solution is correct.

(b) 8 + 4(x – 1) = 0 Let’s pretend the solution obtained was x = 2. Test it:

LHS = 8 + 4(x – 1) = 8 + 4[(2) – 1] = 8 + 4(2 – 1) = 8 + 4(1) = 8 + 4 = 12

RHS = 0

Because the LHS ≠ RHS we know that 2 is not a solution to this equation.

The real solution is, of course,: x = –1. Let’s check it:

LHS = 8 + 4(x – 1) = 8 + 4[(–1) – 1] = 8 + 4(–1 – 1) = 8 + 4(–2) = 8 – 8 = 0

Now the LHS = RHS, and we have confirmed that x = –1 is the correct solution.

(c) x(x + 3) = x2 + 6 solution: x= 2

LHS = x(x + 3) = (2)((2) + 3) = 2(2 + 3) = 2(5) = 10

RHS = x2 + 6 = (2)2 + 6 = 4 + 6 = 10

LHS = RHS, therefore x = 2 is the correct solution.

(d) ½ (4x + 6) = 1 solution: x = –1

LHS = ½ (4x + 6) = ½ (4(–1) + 6) = ½ (–4 + 6) = ½ (2) = 1

RHS = 1

LHS = RHS, therefore x = –1 is the correct solution.

Now go back to problems 5, 6 and 7 and check your answers in the same way.

If we go back to the special cases in question 9, we can check them too:

(a) 2(x + 1) = x + 2 gives a solution: x = 0

LHS = 2(x + 1) = 2((0) + 1) = 2(0 + 1) = 2(1) = 2

RHS = x + 2 = (0) + 2 = 2

LHS = RHS, therefore x = 0 is the correct solution.

(b) 2(x + 3) = 2x + 6 gave a solution of any number! Let’s use 5; you can try another.

LHS = 2(x + 3) = 2((5) + 3) = 2(5 + 3) = 2(8) = 16

RHS = 2x + 6 = 2(5) + 6 = 10 + 6 = 16

LHS = RHS as x = 5. In fact, LHS will equal RHS for any value.

(c) 3 – 2x = –2(1 + x) No number will give a solution; let’s try 12. You can try some more.

LHS = 3 – 2x = 3 – 2(12) = 3 – 24 = – 21

RHS = –2(1 + x) = –2(1 + (12)) = –2(1 + 12) = –2(13) = –26

LHS ≠ RHS and they won’t be equal for any number.

ACTIVITY 4

To tell expressions and equations apart

[LO 2.1, 2.6]

  • Expressions are combinations of letters (a, b, x, y, etc), operations (+, –, ×, ) and numbers (1, –5, π, ½ , etc.) as well as brackets and other signs. An expression does not include equal signs.
  • An expression is a little like a word or a phrase – it does not have a verb.
  • Here are some examples: x, x3, 5½ , 2πr, 5(ab – bc), 5a3 – 3a2 + a – 3, 2a+b2a+b size 12{ sqrt {2 left (a+b right )} } {}, 5a−42a25a−42a2 size 12{ { {5a - 4} over {2a rSup { size 8{2} } } } } {}, etc.
  • An expression can only be manipulated, usually to simplify it. It cannot be solved; it has no solution. The only way to check your work is do the steps backwards, to see whether you come back to the first step.
  • An equation is two expressions with an equal sign in between!
  • It is like a sentence with a verb; it makes a statement. For example 2x – 3 = 45 says that double a certain number, with three subtracted, is equal to 45. Our job is to find that number.
  • Equations must be solved; they have solutions that can be checked.
  • We do a lot of simplifying while we solve equations, but we do more – we are allowed to do more. Remember we could subtract or add terms, as long as we do it to both sides! We could multiply or divide by factors, as long as we do it to both sides. Because an expression does not have two sides we can’t do these things to expressions. Be very careful not to mix them up, and practise until you know instinctively what to do.

ACTIVITY 5

To solve two equations simultaneously

[LO 2.4, 2.9]

Figure 2
Figure 2 (Picture 13.png)

1. The line in figure 1 has defining equation y = 2.

Question: Does the point (1 ; 1) lie on the line?

Answer: We can obtain the answer graphically (by looking at the graph). So we can see that the point does not lie on the line, making the answer no.

We can obtain the answer algebraically, as follows: Substitute the point (1 ; 1) for (x;y) in the equation. Do the LHS and RHS separately as before.

LHS: y = (1) = 2 RHS: 2 LHS ≠ RHS – the point (1 ; 1) does not lie on y = 2.

Question: Does the point (–2 ; 2) lie on the line?

Graphically: Yes.

Algebraically: LHS: y = (2) = 2 RHS: 2 LHS = RHS; yes it does.

Question: Does the point (1½ ; 2) lie on the line? Find the answer both graphically and algebraically.

2. The line in figure 2 is defined by the equation y = 2x – 1.

Questions: Does the point (0 ; 0) lie on the line?

Does the point (1 ; 1) lie on y = 2x – 1?

Does the point (1½ ; 2) lie on the line?

3. In figure 3 the same two lines are drawn together on the same set of axes.

Answer graphically: Which point lies on both lines? The answers to questions 1 and 2 above will be helpful.

It is easy to see from the graph that the only point that lies on both lines is (1½ ; 2).

  • This can also be determined algebraically:

From the line y = 2 we can see that y has the value 2. If we substitute this value into the equation y = 2x – 1.

We can solve the equation to get a value for x. So:

Substitute: (2) = 2x – 1 and solve for x:

2 = 2x – 1 now move the x–terms to the left

–2x + 2 = –1 now move the constant terms to the right

–2x = –2 – 1 simplify

–2x= –3 divide both sides by –2

x = –3  –2 simplify

x = 1½

This shows that the point where the lines cross is (x ; y) = (1½ ; 2).

  • In this method we have solved the two equations simultaneously, to find the values of the variables, which make both equations true. If an equation has only one variable, we need only one equation to find that value of the variable that makes the equation true. If we have two variables, we need two equations to solve for both variables.

Problems:

1 Solve algebraically for a and b: 2a – 3b = 0 and a = 6.

2 Where do the lines y = –x + 5 and y= –1 cross? Find the answer algebraically.

3 Does the point (3 ; 4) lie on both line y = 4 and line y = –x + 1? Do algebraically.

4 Do the lines y = –2 and y = 2 intersect? Find the answer algebraically.

ACTIVITY 6

To solve simple exponential equations

[LO 2.4, 2.8]

Problems and some answers

1 I am thinking of a number that gives 100 when squared. What is the number?

The number could be 10, because 102 = 100. But is –10 not also a correct answer?

Yes, this problem has two valid answers!

Making an equation from this statement means starting by letting x be the number.

x2 = 100

x2 = 102 of x2 = (–10)2 The brackets are essential – can you see that?

x = 10 of x = –10 Both answers are valid.

2 I am thinking of a negative number that gives 25 when squared. What is it?

Let the number be y

y2 = 25

y2 = (5)2 of y2 = (–5)2

y = 5 of y = –5 are the two solutions given by the equation.

According to the problem statement, though, only y = –5 is a valid answer.

3 There is a number that gives 27 when it is cubed. Find the number.

Let the number be x

x3 = 27 x3 = 33x = 3.

Why can’t x be –3?

4 If I cube a certain number I get –8. What is the number?

5 Solve for x, and check your answers by the LHS/RHS method:

a) x2 = 64 b) x2 = 36 c) x2 = –100 d) x2 – 49 = 0

e) x2 = 12,25 f) 3x2 = 12 g) 2x2 – 10,58 = 0

6 Solve for a and check your answers:

a) a3 = 64 b) a3 + 1 = 0 c) 2a2 = 16 d) a4 = 81

Score yourself on the last 12 problems:

Assessment

Table 2
LO 2
Patterns, Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
We know this when the learner:
2.1 investigates, in different ways, a variety of numeric and geometric patterns and relation­ships by representing and generalising them, and by explaining and justifying the rules that generate them (including patterns found in nature and cultural forms and patterns of the learner’s own creation;
2.2 represents and uses relationships between variables in order to determine input and/or output values in a variety of ways using:
2.2.1 verbal descriptions;
2.2.2 flow diagrams;
2.2.3 tables;
2.2.4 formulae and equations;
2.3 constructs mathematical models that repre­sent, describe and provide solutions to pro­blem situations, showing responsibility to­ward the environment and health of others (including problems within human rights, social, economic, cultural and environmental contexts);
2.4 solves equations by inspection, trial-and-improvement or algebraic processes (additive and multiplicative inverses, and factorisa­tion), checking the solution by substitution;
2.5 draws graphs on the Cartesian plane for given equations (in two variables), or deter­mines equations or formulae from given graphs using tables where necessary;
2.6 determines, analyses and interprets the equivalence of different descriptions of the same relationship or rule presented:
2.6.1 verbally;
2.6.2 in flow diagrams;
2.6.3 in tables;
2.6.4 by equations or expressions;
2.6.5 by graphs on the Cartesian plane in order to select the most useful represen­ta­tion for a given situation;
2.8 uses the laws of exponents to simplify expressions and solve equations;
2.9 uses factorisation to simplify algebraic expressions and solve equations.

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