Solving Algebraic Equations

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Algebraic equations are the equations which can consist of different constant term (also called the numerical value), a variable term (usually denoted by alphabets like a, b, x, etc.) which are raised to an exponent or degree of various integers. There are various algebraic properties which help to simplify or evaluate given algebraic equations.

Example 1: Simplify the algebraic equation, 6 (x - 5) + 2 (4x – 1) + 3x2 and evaluate its value when x = 5.

Solution: In order to simplify the above expression, we first use the Distributive Property and multiply the number to the braces.
(6x - 30) + (8 x – 2) + 3x2
Now combine the like terms
6x - 30 + 8x – 2 + 3x2 = 14 x -32 + 3x2
To evaluate the simplified above expression, we plug in the place of ‘x’ as 5.
14 (5) -32 + 3(5)2
70 - 32 + 75 = 113
Hence the solution is 113.
 
Example 2: Simplify the algebraic equation, 2(x - 5) + 4 (4x – 1) + x2 and evaluate its value when x = 1.

Solution: In order to simplify the above expression, we first use the Distributive Property and multiply the number to the braces.
(2x - 10) + (16 x – 4) + x2
Now combine the like terms
2x - 10 + 16 x – 4 + x2 = 18 x - 14 + x2
To evaluate the simplified above expression, we plug in the place of ‘x’ as 1.
18 (1) - 14 + (1)2
18 - 14 + 1 = 5
Hence the solution is 5

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