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Classify Equations: Mini-Lesson Review

Classify Equations: Mini-Lesson Review

Classify Equations: Mini-Lesson Review

Mini Lesson Question

Question #2: Classify Equations

Which of the following correctly classifies the equation and the solution?
6 ( 2 n − 1 ) + 3 = 2 n − 8 + 5 ( 2 n + 1 )
  1. contradiction; no solution
  2. conditional equation; 1 n = − 1 4
  3. conditional equation; n = 0
  4. identity; all real numbers

Use Solutions to Classify Equations

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Consider the equation 7 x + 8 = − 13 7 x + 8 = − 13 .

Subtract 8 to get the constants on one side. 7 x = − 21 7 x = − 21
Divide by 7 to make the coefficient of the variable 1. x = − 3 x = − 3
Steps demonstrating how to solve a linear equation for x, showing the mathematical expression at each stage.

The solution is x = − 3 x = − 3 . This means the equation  7 x + 8 = − 13 7 x + 8 = − 13 is true when we replace the variable, x x , with the value − 3 − 3 , but it is not true when we replace x x with any other value. Whether the equation 7 x + 8 = − 13 7 x + 8 = − 13 is true depends on the value of the variable. The equation is a conditional equation.

An equation that is true for any value of the variable is called an identity. The solution of an identity is all real numbers.

Consider the equation 2 y + 6 = 2 ( y + 3 ) 2 y + 6 = 2 ( y + 3 ) . Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y y .

Distribute. 2 y + 6 = 2 y + 6 2 y + 6 = 2 y + 6
Subtract 2 y 2 y to get the variables on one side. 6 = 6 6 = 6
Steps demonstrating the simplification of an identity equation, showing the distribution and subtraction of terms leading to a true statement.

But 6 = 6 6 = 6 is true. This means the equation  2 y + 6 = 2 ( y + 3 ) 2 y + 6 = 2 ( y + 3 ) is true for any value of y y . The equation is an identity, and we say the solution is all real numbers.

An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.

Consider the equation 5 z = 5 z − 1 5 z = 5 z − 1 .

Subtract 5 z 5 z to get the variables on one side. 0 ≠ − 1 0 ≠ − 1
A mathematical step demonstrating the subtraction of a variable term and its resulting inequality.

The table summarizes the types of equations and solutions.

Type of Equation What happens when you solve it? Solution
Conditional Equation

True for one or more values of the variables and false for all other values

One or more values
Identity

True for any value of the variable

 All real numbers
Contradiction

False for all values of the variable

No solution

>

Try it

Use Solutions to Classify Equations

Classify the following:  5 m + 3 ( 9 + 3 m ) = 2 ( 7 m − 11 ) 5 m + 3 ( 9 + 3 m ) = 2 ( 7 m − 11 ) .

Here is how to classify the equation by finding the solution:

Classify. 5 m + 3 ( 9 + 3 m ) = 2 ( 7 m − 11 ) 5 m + 3 ( 9 + 3 m ) = 2 ( 7 m − 11 )
Distribute. 5 m + 27 + 9 m = 14 m − 22 5 m + 27 + 9 m = 14 m − 22
Combine like terms. 14 m + 27 = 14 m − 22 14 m + 27 = 14 m − 22
Subtract 14 m 14 m from both sides./th> 14 m − 14 m + 27 = 14 m − 14 m − 22 14 m − 14 m + 27 = 14 m − 14 m − 22
Simplify. 27 ≠ − 22 27 ≠ − 22
But 27 ≠ − 22 27 ≠ − 22 The equation is a contradiction. It has no solution.
Illustrates the step-by-step algebraic solution of a linear equation that simplifies to a contradiction, meaning it has no solution.

Check Your Understanding

Which of the following correctly classifies the equation and the solution? 4 + 9 ( 3 x − 7 ) = − 42 x − 13 + 23 ( 3 x − 2 ) 4 + 9 ( 3 x − 7 ) = − 42 x − 13 + 23 ( 3 x − 2 )

  1. identity; all real numbers

  2. conditional equation; x = 0 x = 0

  3. conditional equation; x = 2 5 27 x = 2 5 27

  4. contradiction; no solution

The answer is identity; all real numbers.

Check yourself: Use the Distributive Property on both sides of the equation to eliminate parentheses. After combining like terms, the equation is 27 x − 59 = 27 x − 59 27 x − 59 = 27 x − 59 . Subtract 27x27x from both sides to get − 59 = − 59 − 59 = − 59 . This is a true statement, so the equation is an identity and is true for all values of xx.

Video: Solving Linear Equations

Khan Academy: Number of Solutions to Linear Equations

Watch the video to see how a linear equation may have one solution, no solution, or infinite solutions.

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