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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.

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