3.4 Solving simple problems by forming and solving equations  (Page 2/5)

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

3 x + 90 = 495 simplify

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

3 x = 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 e x perienced 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

2 x – 6 = 28 simplify

2 x = 28 + 6 Add 6 to each side

2 x = 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: 2 x – 6 = 28 2 x = 28 + 6 2 x = 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) 5 x = 35 (b) 4 x = 22 (c) 3 x – 90 = 0 (d) ½ x = 21

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

7 (a) 5 x + 15 = 2 x (b) 8 + 4 x = 22 – 2 x (c) 3 x – 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) = x ² + 6 (d) ½ (4 x + 6) = 1

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