Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Mini Lesson Question

Question #2: Classify Equations

Which of the following correctly classifies the equation and the solution?

6(2n - 1) + 3 = 2n - 8 + 5(2n + 1)

contradiction; no solution
conditional equation; $1n = - \frac{1}{4}$
conditional equation; $n = 0$
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 $7 x + 8 = - 13$ .

Subtract 8 to get the constants on one side.	$7 x = - 21$
Divide by 7 to make the coefficient of the variable 1.	$x = - 3$

The solution is $x = - 3$ . This means the equation $7 x + 8 = - 13$ is true when we replace the variable, $x$ , with the value $- 3$ , but it is not true when we replace $x$ with any other value. Whether the equation $7 x + 8 = - 13$ 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 $2 y + 6 = 2 (y + 3)$ . Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for $y$ .

Distribute.	$2 y + 6 = 2 y + 6$
Subtract $2 y$ to get the variables on one side.	$6 = 6$

But $6 = 6$ is true. This means the equation $2 y + 6 = 2 (y + 3)$ is true for any value of $y$ . 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 $5 z = 5 z - 1$ .

Subtract $5 z$ to get the variables on one side.

0 \neq - 1

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

Try It: Use Solutions to Classify Equations

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

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

Classify.	$5 m + 3 (9 + 3 m) = 2 (7 m - 11)$
Distribute.	$5 m + 27 + 9 m = 14 m - 22$
Combine like terms.	$14 m + 27 = 14 m - 22$
Subtract $14 m$ from both sides./th>	$14 m - 14 m + 27 = 14 m - 14 m - 22$
Simplify.	$27 \neq - 22$
But $27 \neq - 22$	The equation is a contradiction. It has no solution.

Check Your Understanding

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

Multiple Choice:

identity; all real numbers
conditional equation; $x = 0$
conditional equation; $x = 2 \frac{5}{27}$
contradiction; no solution

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 $27 x - 59 = 27 x - 59$ . Subtract $27 x$ from both sides to get $- 59 = 59$ . This is a true statement, so the equation is an identity and is true for all values of $x$ .

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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Classify Equations: Mini-Lesson Review