3.4 Solving simple problems by forming and solving equations  (Page 4/5)

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 , x ³ , 5½ , 2πr, 5( ab – bc ), 5 a ³ – 3 a ² + a – 3, $\sqrt{2\left(a+b\right)}$ , $\frac{\mathrm{5a}-4}{{\mathrm{2a}}^2}$ , 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 e x pressions with an equal sign in between!
• It is like a sentence with a verb; it makes a statement. For example 2 x – 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]

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 = 2 x – 1.

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

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

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

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

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

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

–2 x = –2 – 1 simplify

–2 x = –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).