HOW TO EVALUATE LINEAR EXPRESSIONS
To evaluate a linear expression, replace the variable or variables by the given values and simplify using order of operations or 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.
Evaluate the following linear expressions using the given values of the variables.
Example 1 :
16x - 48 for x = -1
Solution :
= 16x - 48
Substitute x = -1.
= 16(-1) - 48
= -16 - 48
= -64
Example 2 :
-11x - 23 for x = 5
Solution :
-11x - 23
Substitute x = 5.
= -11(5) - 23
= -55 - 23
= -78
Example 3 :
14x + 99 for x = 0
Solution :
= 14x + 99
Substitute x = 0.
= 14(0) + 99
= 0 + 99
= 99
Example 4 :
-7y + 69 for y = 9
Solution :
= -7y + 69
Substitute y = 9.
= -7(9) + 69
= -63 + 69
= 6
Example 5 :
-10t + 35 for t = 5
Solution :
= -10t + 35
Substitute t = 5.
= -10(5) + 35
= -50 + 35
= -15
Example 6 :
18x - 32 for x = -3
Solution :
= 18x - 32
Substitute x = -3.
= 18(-3) - 32
= -54 -32
= -86
Example 7 :
-4x + 7 for x = -1
Solution :
= -4x + 7
Substitute x = -1.
= -4(-1) + 7
= 4 + 7
= 11
Example 8 :
-5x + 5 for x = 1
Solution :
= -5x + 5
Substitute x = 1.
= -5(1) + 5
= -5 + 5
= 0
Example 9 :
9 (4x + 2) for x = -3
Solution :
= 9(4x + 2)
Substitute x = -3.
= 9[4(-3) + 2]
= 9[-12 + 2]
= 9(-10)
= -90
Example 10 :
2(b - 6) for b = 5
Solution :
= 2(b - 6)
Substitute b = 5.
= 2(5 - 6)
= 2(-1)
= -2
Example 11 :
(y + x) ÷ 2 + x for x = 1 and y = 1
Solution :
= (y + x) ÷ 2 + x
Substitute x = 1 and y = 1.
= (1 + 1) ÷ 2 + 1
= 2 ÷ 2 + 1
= 1 + 1
= 2
Example 12 :
z - (y ÷ 3 - 1) for y = 3, and z = 7
Solution :
= z - (y ÷ 3 - 1)
Substitute y = 3 and z = 7.
= 7 - (3 ÷ 3 - 1)
= 7 - (1 - 1)
= 7 - 0
= 7
Example 13 :
p - (9 - (m + q)) for m = 4, p = 5 and q = 3
Solution :
= p - (9 - (m + q))
Substitute m = 4, p = 5 and q = 3.
= 5 - (9 - (4 + 3))
= 5 - (9 - 7)
= 5 - 2
= 3
Example 14 :
y - (4 - x - y ÷ 2) for x = 3, and y = 2
Solution :
= y - (4 - x - y ÷ 2)
Substitute x = 3 and y = 2.
= 2 - (4 - 3 - 2 ÷ 2)
= 2 - (4 - 3 - 1)
= 2 - (4 - 4)
= 2 - 0
= 2
Example 15 :
y ÷ 5 + 1 + x ÷ 6 for x = 6, and y = 5
Solution :
= y ÷ 5 + 1 + x ÷ 6
Substitute x = 6 and y = 5.
= 5 ÷ 5 + 1 + 6 ÷ 6
= 1 + 1 + 1
= 3
Example 16 :
x - (5 - 2(y + z)) for x = 4, y = 5, and z = 3
Solution :
= x - (5 - 2(y + z))
Substitute x = 4, y = 5 and z = 3.
= 4 - (5 - 2(5 + 3))
= 4 - (5 - 2(8))
= 4 - (5 - 16)
= 4 - (-11)
= 4 + 11
= 15
Example 17 :
6q + (2m + 1) ÷ 17 for m = 8, and q = 3
Solution :
= 6q + (2m + 1) ÷ 17
Substitute m = 8 and q = 3.
= 6(3) + (2(8) + 1) ÷ 17
= 18 + (16 + 1) ÷ 17
= 18 + 17 ÷ 17
= 18 + 1
= 19
Problem 18 :
a3 - (b2 + c) ÷ a + (ab + c) for a = 4, b = -3 and c = 7
Solution :
a3 - (b2 + c) ÷ a + (ab + c)
Substitute a = 4, b = -3 and c = 7.
= 43 - [(-3)2 + 7] ÷ 4 + [4(-3) + 7]
= 43 - [(-3)2 + 7] ÷ 4 + [4(-3) + 7]
= 43 - [9 + 7] ÷ 4 + [4(-3) + 7]
= 43 - 16 ÷ 4 + [-12 + 7]
= 43 - 16 ÷ 4 - 5
= 64 - 16 ÷ 4 - 5
= 64 - 4 - 5
= 60 - 5
= 55
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