Algebra plays a major role in mathematics that deals with solving the algebraic expressions in order to calculate the unknown variable value. Algebra mainly involves calculating the unknown variable value with the help of known values. In algebra alphabets are used to represents variables and the integers are considered as constants. The following are the example problem which helps to know the basic things in algebra. The solved problems are discussed below with detailed solutions.

**Example 1:** When z <3, solve

|z - 3| - 4|-6|

**Solution:**

Given expression is,

|z - 3| - 4|-6|

When z < 3 then z - 3 < 3 and if z - 3 < 3 the |z - 3| = -(z - 3).

Now substitute |z - 3| by -(z - 3) and |-6| by 6

|z - 3| - 4|-6| = -(z - 3) -4(6)

**|z - 3| - 4|-6| = -z -21** **is the solution.**

**Example 2:** Solve the algebraic equation

-3a - 2 - (a - 3) = -4(4a + 5) + 13

**Solution:**

Given algebraic expression is

(-3a - 2) - (a - 3) = -4(4a + 5) + 13

Multiplying the above factors

-3a - 2 - a + 3 = -16a - 20 +13

Now combining the above terms we get

-4a - 1 = -16a - 7

Add 16a + 7 on both sides of the above expression then the equation becomes,

**12a = - 6**

**a = - 1/ 2 is the solution.**

**Example 3:** Simplify the algebraic expression

2(x -3) + 4z - 2(x -z -3)

**Solution:**

Given the algebraic expression

2(x -3) + 4z - 2(x -z -3)

Multiplying the integer terms

= 2x - 6 + 4z -2x + 2z + 6

Now grouping the above terms we get

= **6z is the solution**

1) Solve the algebraic equation -2(s - 1) – 4s - 1 = 3(s + 2)

**Answer:** **s = - 5/9 is the solution.**

2) Solve the algebraic equation -4(s + 2) = s + 5

**Answer:** **s = -13/5 is the solution.**