# Reducing Equations to Simpler Form

Reducing Equations to Simpler form0 %

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## Example 3

**Solve**

- Multiplying both sides of the equation by 6
- On simplifying the divisions
- Solving for terms within the brackets
- Solving the constant terms on LHS.
- Move x to one side of the equation to simplify.
- Simplify for the
*x*vaiable - Now, simplify the constant terms
- Solve the constant terms on the RHS
- x =
- We have found the answer

Now, that we have found the solution, let's check if the LHS and RHS of the equation are equal or not.

Check LHS and RHS0 %

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**Now check the LHS and RHS of the above expression by substituting the obtained x value.**

LHS = 6(-1) + 1 3 + 1 = -6+1 3 + 1 = − 5 3 + 3 3 = -5+3 3 =

RHS = (-1)-3 6 = -4 6 =

LHS = RHS. (as required)

Solve the expression0 %

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## Example 4

**Solve 5x – 2 (2x – 7) = 2 (3x – 1) + **

- Solving the terms with brackets
- This can be further simplified
- Solving for the constant terms on RHS.
- Transposing the constant to the LHS and
*x*to RHS - Solving we get
- Now, we have
- Therefore, required solution is x =
- We have found the answer.

**Check:** LHS = 5 ×

=

RHS = 2(

=