About No Solution Equation:

Once we have mastered solving a linear equation in one variable, we begin to consider the types of equations that we will encounter. We will see conditional equations, identities, and contradictions. Our two special case scenarios are identities (equations with infinitely many solutions) and contradictions (no solution equation).


Test Objectives
  • Demonstrate the ability to solve a conditional equation
  • Demonstrate an understanding of a contradiction
  • Demonstrate an understanding of an identity
No Solution Equation Practice Test:

#1:

Instructions: Identify each equation as conditional, an identity, or a contradiction.

$$a)\hspace{.25em} 6(1 - 3k) = -6(12 + 3k)$$

$$a)\hspace{.25em} 6(1 - 3k) =$$$$ -6(12 + 3k)$$


#2:

Instructions: Identify each equation as conditional, an identity, or a contradiction.

$$a)\hspace{.25em} -2b + 10(1 + b) = -2(-2b - 8) - 6$$

$$a)\hspace{.25em} -2b + 10(1 + b) =$$$$ -2(-2b - 8) - 6$$


#3:

Instructions: Identify each equation as conditional, an identity, or a contradiction.

$$a)\hspace{.25em} 115 - 10p = -12 - 5(2p - 6)$$

$$a)\hspace{.25em} 115 - 10p =$$$$ -12 - 5(2p - 6)$$


#4:

Instructions: Identify each equation as conditional, an identity, or a contradiction.

$$a)\hspace{.25em}-11(r+9)=-6r+5\left(-r-\frac{99}{5}\right)$$

$$a)\hspace{.25em}-11(r+9)=$$$$-6r+5\left(-r-\frac{99}{5}\right)$$


#5:

Instructions: Identify each equation as conditional, an identity, or a contradiction.

$$a)\hspace{.25em} -9(1 + 5x) = -4(11x - 3)$$

$$a)\hspace{.25em} -9(1 + 5x) =$$$$ -4(11x - 3)$$


Written Solutions:

#1:

Solutions:

a) contradiction

a) contradiction


#2:

Solutions:

a) conditional

a) conditional


#3:

Solutions:

a) contradiction

a) contradiction


#4:

Solutions:

a) identity

a) identity


#5:

Solutions:

a) conditional

a) conditional