Algebraic expressions and equations normally come in integer or fraction form.
1) 4x - 3 = x
2) (3/4)x - 3x = 1/(3x)
Simplification of the above examples will not pose much of a problem except maybe in the challenge of bringing the numbers and unknowns over the "equal" sign.
But algebraic equations can come in decimal form too.
0.3(0.2x - 1) = 0.1x
How do we go about solving the above "decimated" algebraic equation easily?
A simple trick that I can think of (or maybe too simply a technique to call it 'trick").
What I would do is to multiply the expression on both sides by 10.
The idea is to bring the decimal number into the integer range.
BUT do note that the expression on the left side has two decimal numbers.
As such I would have to "x 10" twice.
This means that there is a "x 100" on the left and right side.
The new equation will thus be:
3 (2x - 10) = 10 x
==> 6x - 30 = 10x
==> -30 = 10x - 6x = 4x
==> x = -30 / 4 = -7.5
Decimal can be seen to be intimidating when in the decimal form. However, it can be elevated to the familiar integer form through simple multiplication.
However, do take note of how many decimal number has been multiplied.
Left and right sides of the equation has to have the same number of multiplication (or division) to stay equal and valid.
Maths is not that frightening.
It can be interesting, if the method to "attack" it is properly done.