3.4 Solving simple problems by forming and solving equations  (Page 3/5)

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

(b) 2 x + 6 = 2 x + 6 2 x – 2 x = 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 – 2 x = –2 – 2 x –2 x + 2 x = –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) = x ² + 6 solution: x = 2

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

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

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

(d) ½ (4 x + 6) = 1 solution: x = –1

LHS = ½ (4 x + 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) = 2 x + 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 = 2 x + 6 = 2( 5 ) + 6 = 10 + 6 = 16

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

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

LHS = 3 – 2 x = 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.