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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 36 Page 106

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side. Start by using the Distributive Property.

a=13/33

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other side using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the right-hand side of the equation.

3a+1=- 3.6(a-1)

Distribute - 3.6

3a+1=-3.6a-(-3.6)

a-(- b)=a+b

3a+1=-3.6a+3.6

Now we can continue to solve using the Properties of Equality.

3a+1=-3.6a+3.6

LHS+3.6a=RHS+3.6a

3a+1+3.6a=-3.6a+3.6+3.6a

Add terms

6.6a+1=3.6

LHS-1=RHS-1

6.6a+1-1=3.6-1

Subtract terms

6.6a=2.6

.LHS /6.6.=.RHS /6.6.

6.6a/6.6=2.6/6.6

Calculate quotient

a=2.6/6.6

a/b=a * 5/b * 5

a=13/33

The equation has one solution, a= 1333.

Subchapter links

Lesson Check p.105

Practice and Problem-Solving Exercises p.105-108

Standardized Test Prep p.108

Mixed Review p.108