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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

4. Solving Equations With Variables on Both Sides

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Exercise 27 Page 106

Simplify the equation as much as possible to determine the number of solutions. Use the Properties of Equality and Distributive Property.

No solution

Practice makes perfect

When identifying the number of solutions there are to an equation, there are three possible outcomes.

Result of Simplifying	Number of Solutions
Variable on one side and constant on the other	One solution
Identity — both sides are identical	All real numbers
Contradiction — both sides are different numbers	No solutions

To do find the number of solutions, we will simplify both sides of the given equation as much as possible and observe the result. Let's start by using the Distributive Property.

2(a-4)=4a-(2a+4)

Distribute 2

2a-8=4a-(2a+4)

Distribute -1

2a-8=4a-2a-4

Subtract term

2a-8=2a-4

LHS-2a=RHS-2a

2a-8-2a=2a-4-2a

Subtract terms

- 8 = - 4 *

Simplifying the equation resulted in a contradiction. Therefore, the equation has no solution.

Subchapter links

Lesson Check p.105

Practice and Problem-Solving Exercises p.105-108

Standardized Test Prep p.108

Mixed Review p.108