Solving Linear Equations
The two examples below show that it is not true that the solution to every
linear equation is a unique real number.
Example 1
Solve: 4(x  2) = 10x
+ 5  6x
Solution 
4(x  2) 
= 10x
+ 5  6x 
Step 1 Remove parentheses.
Distribute the 4.

4x  8 
= 10x + 5  6x 
Step 2 On each side of the equation, combine like
terms.
Combine 10x and 6x.

4x  8

= 4x + 5 
Step 3 Isolate the variable.
Subtract 4x from both sides.
Simplify. 
4x  8  4x
8 
= 4x + 5  4x
= 5 
The result, 8 = 5 is a not a
true statement. So, there is no value of x which will make the original equation
true.
Therefore, the equation 4(x  2) =
10x
+ 5  6x has no solution.
Note:
The statement 8 = 5 is a false statement,
which is sometimes called a contradiction.
Example 2
Solve: 3(y  2) + 2y
= 6 + 5y
Solution 
3(y  2)
+ 2y 
= 6 + 5y 
Step 1 Remove parentheses.
Distribute the 3.

3y  6 + 2y 
= 6 + 5y 
Step 2 On each side of the equation,
combine like terms.
Combine 3y and 2y.

5y  6 
= 6 + 5y 
Step 3 Isolate the variable.
Subtract 5y from both sides.
Simplify. 
5y  6  5y
6 
= 6 + 5y  5y
6 
The result, 6 = 6 is a true statement. Therefore, any value of y will
make the original equation true.
So, the solution of 3(y  2) + 2y
= 6 + 5y is all real numbers.
Note:
Here is another way to solve
5y  6 = 6 + 5y.
Add 6.
Divide by 5. 
5y
y 
= 5y
= y 
The statement y = y is true for all real
numbers.
The statement  6 = 6 is a true
statement, which is sometimes called an identity.
