Equivalent Equations, Transformation Rules
p.1. Concept of Equivalent Equations
Equivalent equations are those that have the same roots.
Equations with no roots are also considered equivalent.
2x + 5 = 7
3x + 6 = 9
x = 1
x = 1
Each of the equations has the same root x=1
$\implies$ the equations are equivalent
(x - 3)(x + 2) = 0
2x + 4 = 0
$x_1 = 3 and x_2 = -2$
x = -2
The first equation has two roots, while the second equation has only one root
$\implies$ the equations are not equivalent
x^2 + 1 = 0
2x^2 + 7 = 0
$\varnothing$
$\varnothing$
Both of the equations have no roots
$\implies$ the equations are equivalent
p.2. Rules of Equivalent Transformations
When solving an equation, try to replace it with a simpler equivalent equation.
For this purpose, use the following rules.
Rules of Equivalent Transformations
- 1. In any part of the equation, it is allowed to expand brackets and combine like terms.
- 2. Any term can be moved from one part of the equation to another by changing its sign.
- 3. Any term can be moved from one part of the equation to another by changing its sign.
As a result of these transformations, we always obtain an equation equivalent to the given one.
p.3. Examples of Problem Solving
Example 1. Solve the equation $ \frac {1}{5}x = 12 - 7x$
Solution:
$ \frac {1}{5}x = 12 - 7x \iff \frac {1}{5}x + 7x = 12 \iff 7 \frac {1}{5}x = 12 \iff x = 12:7 \frac {1}{5} \iff$
$ x = 12 \cdot \frac {5}{36} = \frac {5}{3} =1 \frac {2}{3} $
Answer: x = 1 \frac {2}{3}
Example 2. Solve the equation $ \frac {3x}{7} - \frac {x}{14} = 10$
Solution:
$ \frac {3x}{7} - \frac {x}{14} = 10 | \times 14 \iff 6x - x = 140 \iff 5x = 140 \iff x = 140 : 5 = 28$
Answer: x = 28
Example 3. Solve the equation $7x - \frac {2}{5} =\frac 15 (3x+14)$
Solution:
$7x - \frac 25 = \frac 15 (3x + 14) | \times 5 \iff 35x - 2 = 3x + 14 \iff 35x - 3x = 14 + 2 \iff$
$ \iff 32x = 16 \iff x = \frac {16}{32} = \frac 12$
Answer: x = \frac 12
Example 4. Solve the equation $\frac {5x-1}{2} - \frac {3x+4}{8} = \frac {x-3}{4}$
Solution:
$\frac {5x-1}{2} - \frac {3x+4}{8} = \frac {x-3}{4} | \times 8 \iff 4(5x-1)-(3x+4)=2(x-3) \iff $
$ \iff 15x=2 \iff x= \frac {2}{15} $
Answer: x = $\frac {2}{15}$
Example 5. Find the values of the parameter a such that the equations
3(x-1)=5-x and ax=x+a are equivalent.
Solution:
Find the root of the first equation
$3(x-1)=5-x \iff 3x-3=5-x \iff 3x+x=5+3 \iff 4x=8 \iff x=2$
Substitute this root into the second equation
$a \cdot 2=2+a \iff 2a-a=2 \iff a=2$
For a=2, both of the equations have the same root, x=2.
Answer: a=2
