When we work with equations, we will see three types: conditional equations, contradictions, and identities. Most often, we will deal with conditional equations. These equations are true under certain circumstances and false under all others. Additionally, we will encounter contradictions and identities. Contradictions are equations that don't have a solution, while identities are always true and are said to have an infinite number of solutions.

Test Objectives
• Demonstrate the ability to solve a linear equation in one variable
• Demonstrate the ability to identify an equation as: conditional, an identity, or a contradiction
Types of Equations Practice Test:

#1:

Instructions: Solve each equation and state the equation type.

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$$a)\hspace{.2em}-4(x - 6) + 11(x - 6)=-2x + 12x$$

$$b)\hspace{.2em}-2(6x + 6) + 11=-(1 + 12x)$$

#2:

Instructions: Solve each equation and state the equation type.

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$$a)\hspace{.2em}-12x + 8x=-8(x - 5) - 10(1 - x)$$

$$b)\hspace{.2em}3x + 11(x - 8)=2(x - 2)$$

#3:

Instructions: Solve each equation and state the equation type.

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$$a)\hspace{.2em}-8 - 5(x + 9)=-3(x - 8) - 2x$$

$$b)\hspace{.2em}7(x + 1)=-1 + 7(x - 2)$$

#4:

Instructions: Solve each equation and state the equation type.

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$$a)\hspace{.2em}-\left(3x + \frac{1}{3}\right) - \frac{2}{3}\left(\frac{5}{2}x + \frac{3}{2}\right)=\frac{4}{3}x + \frac{1}{3}x$$

$$b)\hspace{.2em}-\frac{7}{2}- \frac{1}{3}\left(x - \frac{7}{2}\right)=-\frac{1}{3}\left(x + \frac{3}{2}\right)$$

#5:

Instructions: Solve each equation and state the equation type.

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$$a)\hspace{.2em}-2.4(0.6 + x)=-3(0.95 + 0.8x) - 2.5$$

$$b)\hspace{.2em}0.2(x - 2.1)=1.3(1.1 - x) + 1.1x$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x=-14, \hspace{.2em}conditional$$

$$b)\hspace{.2em}(-\infty, \infty), \hspace{.2em}identity$$

#2:

Solutions:

$$a)\hspace{.2em}x=-5, \hspace{.2em}conditional$$

$$b)\hspace{.2em}x=7, \hspace{.2em}conditional$$

#3:

Solutions:

$$a)\hspace{.2em}∅,\hspace{.2em}contradiction$$

$$b)\hspace{.2em}∅,\hspace{.2em}contradiction$$

#4:

Solutions:

$$a)\hspace{.2em}x=-\frac{4}{19}, \hspace{.2em}conditional$$

$$b)\hspace{.2em}∅,\hspace{.2em}contradiction$$

#5:

Solutions:

$$a)\hspace{.2em}∅,\hspace{.2em}contradiction$$

$$b)\hspace{.2em}x=4.625, \hspace{.2em}conditional$$