Exercise 102 Page 153

Practice makes perfect

a Let's see what happens if we substitute 4 for x.

x+2/x-4

Substitute

x= 4

4+2/4-4

SubTerm

Subtract term

4+2/0

We get 0 in the denominator when x=4. Since division by 0 is undefined, x=4 will give an undefined value. Therefore, x cannot be 4.

b We have two expressions.

Part A:& x/3x+1+2x^2-2/(x-5)(3x+1) [1em] Part B:& 9-3x/(x+3)(x-3)+2x/x+3 Let's look at them one at the time.

Part A

The expression is undefined if any of the denominators are 0. 3x+1&=0 or (x-5)(3x+1)&=0 We can solve the equations separately.

3x+1=0

SubEqn

LHS-1=RHS-1

3x=-1

DivEqn

.LHS /3.=.RHS /3.

x=-1/3

Next, we will solve the second equation using the Zero Product Property.

(x-5)(3x+1)=0

ZeroProdProp

Use the Zero Product Property

lcx-5=0 & (I) 3x+1=0 & (II)

▼

(I), (II): Solve for x

AddEqn

(I):LHS+5=RHS+5

lx=5 3x+1=0

SubEqn

(II):LHS-1=RHS-1

lx=5 3x=-1

DivEqn

(II):.LHS /3.=.RHS /3.

lx_1=5 x_2=- 13

In conclusion, there are two x-values for which the expression is undefined. x=5 and x=-1/3

Part B

Same as for the previous expression, we want to find for which x-values any of the denominators are zero.

(x+3)(x-3)=0

ZeroProdProp

Use the Zero Product Property

lcx+3=0 & (I) x-3=0 & (II)

SubEqn

(I):LHS-3=RHS-3

lx=-3 x-3=0

AddEqn

(II):LHS+3=RHS+3

lx_1=-3 x_2=3

Now we solve the other equation. x+3=0 ⇔ x=-3 In conclusion, x=-3 and x=3 are excluded values.

c To create an expression where x=6 and x= 13 are excluded values we need to make sure that when these values are substituted, the denominator is zero. Therefore, the expression must be a fraction. Consider the following fraction.

1/x-6 If x=6, the denominator is zero, so x=6 is an excluded value here. To make x= 13 an excluded value we can add another fraction or simply add the factor (x- 13) in the denominator to the existing fraction. 1/(x-6)(x- 13) If x is either 6 or 13 the denominator is zero, and they are therefore excluded values.