EVALUATE NON LINEAR EXPRESSIONS

To evaluate a nonlinear expression, we have to substitute the given values of the variables into the expression and simplify using PEMDAS.

PEMDAS is the rule that can be used to simplify or evaluate complicated numerical expressions with more than one binary operation.

----> Parentheses

----> Exponent 

M ----> Multiply

----> Divide

----> Add

----> Subtract

Important Notes :

1. In a particular simplification, if you have both multiplication and division, do the operations one by one in the order from left to right.

2. Division does not always come before multiplication. We have to do one by one in the order from left to right.

Example 1 :

Evaluate the following expression for x = 3 :

2x2

Solution :

= 2x2

Substitute x = 3.

= 2(3)2

= 2(3)2

= 2(9)

= 18

1Exponent

Multiplication

Result

Example 2 :

Evaluate the following expression for x = 2 and y = 3 :

x2 + y2

Solution :

= x+ y2

Substitute x = 2 and y = 3.

= 2+ 32

= 2+ 32

4 + 9

= 13

1Exponent

Addition

Result

Example 3 :

Evaluate the following expression for x = -2 :

3x3

Solution :

3x3

Substitute x = -2.

= 3(-2)3

= 3(-2)3

= 3(-8)

= -24

1Exponent

Multiplication

Result

Example 4 :

Evaluate the following expression for x = -1 and y = 2 :

x+ 3y3

Solution :

x+ 3y3

Substitute x = -1 and y = 2.

= (-1)+ 3(2)2

= (-1)+ 3(2)2

= 1 + 3(4)

= 1 + 12

= 13

1Exponent

Multiplication

Addition

Result

Example 5 :

Evaluate the following expression for x = 1 and y = -2 :

3x- 2y3

Solution :

3x- 2y3

Substitute x = 1 and y = -2.

= 3(1)- 2(-2)3

= 3(1)- 2(-2)3

= 3(1) - 2(-8)

= 3 + 16

= 19

1Exponent

Multiplication

Addition

Result

Example 6 :

Evaluate the following expression for x = 1 and y = 2 :

y2 ÷ 2 + x3

Solution :

= y2 ÷ 2 + x3

Substitute x = 1 and y = 2.

= 22 ÷ 2 + 13

= 22 ÷ 2 + 13

= 4 ÷ 2 + 1

2 + 1

= 3

1Exponent

Division

Addition

Result

Example 7 :

Evaluate the following expression for x = 3 and y = -1 :

x2 + y + y2

Solution :

= x2 + y + y2

Substitute x = 3 and y = -1.

= 32 + (-1) + (-1)2

= 32 + (-1) + (-1)2

= 9 - 1 + 1

= 8 + 1

= 9

1Exponent

Subtraction

Addition

Result

Example 8 :

Evaluate the following expression for x = 6, y = 8 and z = 6.

z2(x + y)

Solution :

= z2(x + y)

Substitute x = 6, y = 8 and z = 6.

= 62(6 + 8)

= 62(6 + 8)

= 62(14)

= 36(14)

= 504

1Parentheses

1Exponent

Multiplication

Result

Example 9 :

Evaluate the following expression for x = 1 and y = 1.

(y3 + x) ÷ 2 + x

Solution :

= (y3 + x) ÷ 2 + x

Substitute x = 1 and y = 1.

= (13 + 1) ÷ 2 + 1

= (13 + 1) ÷ 2 + 1

= (13 + 1) ÷ 2 + 1

= (1 + 1) ÷ 2 + 1

= 2 ÷ 2 + 1

= 1 + 1

= 2

1Parentheses

1Exponent

Parentheses

Division

Addition

Result

Example 10 :

Evaluate the following expression for y = 3 and z = 7.

z3 - (y ÷ 3 - 1)

Solution :

= z3 - (y ÷ 3 - 1)

Substitute y = 3 and z = 7.

= 73 - (3 ÷ 3 - 1)

= 73 - (3 ÷ 3 - 1)

= 73 - (3 ÷ 3 - 1)

= 73 - (1 - 1)

= 73 - 0

= 243

1Parentheses

1Division

1Subtraction

1Exponent

Result

Example 11 :

You are saving for a skateboard. Your aunt gives you $45 to start and you save $3 each week.

evaluating-non-linear-exp-q1

The expression 45 + 3w gives the amount of money you save after w weeks.

a. How much will you have after 4 weeks, 10 weeks, and 20 weeks?

b. After 20 weeks, can you buy the skateboard? Explain.

Solution :

a)

Given expression is,

saving after w weeks = 45 + 3w

When w = 4

Savings after 4 weeks = 45 + 3(4)

= 45 + 12

= 57

When w = 10

Savings after 10 weeks = 45 + 3(10)

= 45 + 30

= 75

When w = 20

Savings after 20 weeks = 45 + 3(20)

= 45 + 60

= 105

b) The amount we have after 20 weeks is $105. The price of the skate board is $125. Then, the amount is not enough.

Example 12 :

Write and evaluate an expression for the problem.

a) You receive $8 for raking leaves for 2 hours. What is your hourly wage?

b)  Music lessons cost $20 per week. How much do 6 weeks of lessons cost?

c) The scores on your first two history tests were 82 and 95. By how many points did you improve on your second test?

Solution :

a)

Let x be the wage for one hour.

Amount recived for 2 x = $8

x = 8/2

Hourly wage = $4

b) 

Cost per week = 20

Number of weeks of work = 6

So, cost of lessons = 6(20)

= 120

c)

Score for first test = 82

Score for second test = 95

Points to be improved = 95 - 82

= 13

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