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.
P ----> Parentheses
E ----> Exponent
M ----> Multiply
D ----> Divide
A ----> Add
S ----> 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 :
= x2 + y2
Substitute x = 2 and y = 3.
= 22 + 32
= 22 + 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 :
x2 + 3y3
Solution :
= x2 + 3y3
Substitute x = -1 and y = 2.
= (-1)2 + 3(2)2
= (-1)2 + 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 :
3x2 - 2y3
Solution :
= 3x2 - 2y3
Substitute x = 1 and y = -2.
= 3(1)2 - 2(-2)3
= 3(1)2 - 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.
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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