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Chapter 3

1st Degree Polynomial Equations
ax+b=0

 
3.2. More Examples

Example 1
Solve the equation: 
Answer

When the equation is an equation of two fractions like above, then we can eliminate the denominators by multiplying the nominator of each fraction with the denominator of the other.

<=> 3 (2x +1)=4 ( x - 1) <=> 6x + 3=4x - 4 <=> 6x - 4x = -3 - 4 <=> 2x = -7 <=>

 
 <=> x = -7/2

Example 2

Solve the equation: 

Answer

Here we follow the algorithm that we described in the previous paragraph.

  <=>  <=> 3(5x+1) - x = 2(7x+4) <=>
<=> 15x + 3 - x = 14x + 8 <=> 15x - x - 14x = -3 + 8 <=> (15 - 1 - 14)x = 5 <=>
 
 <=> 0x = 5 which has no solutions in R
 

Example 3

Solve the equation: 

Answer

  <=>  <=>

2(2x - 3) - (3x + 1) = x - 3 - 4 <=> 4x - 6 -3x - 1 = x - 3 - 4 <=>

(4 - 3 -1)x = 0 <=>
 
<=> 0x = 0 
  which is true for every