Mathematics Magazine for Grades 1-12  

Grade 8

Theory:

To solve an equation means to find a value for the variable that makes the equation true. Whatever you do to one side of the equation, you must also do to the other side.

Variable
A letter used to represent a number value in an expression or an equation.

To solve the equation we must try to get the variable x alone on one side. We can use the inverse of adding 15 - or subtracting 15 - to get x alone on the left side. Now we have x alone on the left side, since 15 – 15 = 0, but the scale is not in balance. To balance the scale, we must also subtract 15 from the right side of the equation.

x + 15 - 15 = 30 - 15
x = 15

30 – 15 = 15, so we find that x = 15.

We can check this solution by substituting the value 15 for x in the original equation. When we evaluate for x = 15 we get 30 = 30, which is a true statement. We know our solution is correct!

 x + 15 = 30
 (15) + 15 = 30
 30 = 30

Solutions from the Previous Issue:

Calculate:

1.        12 ÷ 4 + 3² =

Solution:

12 ÷ 4 + 3² = 3 + 9 = 12

2.        (4² + 5) – 3 =

Solution:

(4² + 5) – 3 = (16 + 5 ) – 3 = 21  - 3 = 18

3.        20 ÷ (12 - 2) x 3² – 2 =

Solution:

20 ÷ (12 - 2) x 3² – 2 = 20 ÷ 10 x 9 – 2 = 2 x 9 – 2 =16

4.        (2⁰ + 2 ∙ 2⁴) : 11 =

Solution:

(2⁰ + 2 ∙ 2⁴) : 11 = (1 + 2 ∙ 16) : 11 = (1 + 32) : 11 = 3

5.        (5⁰ + 5² ∙ 5) : 63 =

Solution:

(5⁰ + 5² ∙ 5) : 63 = (1 +125) : 63 = 2

6.        (3³∙3 - 3⁰) : 40 =

Solution:

(3³∙3 - 3⁰) : 40 = (81 – 1) : 40 = 80 : 40 = 2

Proposed Exercises:

Calculate:

      4.