EVALUATE LINEAR EXPRESSIONS

Evaluate the following linear expressions using the given values of the variables.

Example 1 :

y ÷ 2 + x; x = 1 and y = 2

Solution :

= y ÷ 2 + x

Substitute x = 1 and y = 2.

= 2 ÷ 2 + 1

= 1 + 1

= 2

Example 2 :

x + 2y; x = 9 and y = 10

Solution :

= x + 2y

Substitute x = 9 and y = 10.

= 9 + 2(10)

= 9 + 20

= 29

Example 3 :

z(x + y); x = 6, y = 8 and z = 6

Solution :

= z(x + y)

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

= 6(6 + 8)

= 6(14)

= 84

Example 4 :

(y + x) ÷ 2 + x; 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 5 :

z - (y ÷ 3 - 1); 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 6 :

p - (9 - (m + q)); 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 7 :

 y - (4 - x - y ÷ 2); 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 8 :

y ÷ 5 + 1 + x ÷ 6; 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 9 :

x - (5 - 2(y + z)); 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 10 :

6q + (2m + 1) ÷ 17; 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

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