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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

2. Standard Form of a Quadratic Function

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Exercise 58 Page 208

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

- 2.5

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, using the Properties of Equality. In this case, we will start by using the Distributive Property to simplify the left-hand side of the equation.

3(a+4)+2(a-1)=a

Distribute 3

3a+12+2(a-1)=a

Distribute 2

3a+12+2a-2=a

Add and subtract terms

5a+10=a

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

5a+10=a

LHS-a=RHS-a

5a+10-a=a-a

Subtract terms

4a+10=0

LHS-10=RHS-10

4a+10-10=0-10

Subtract terms

4a=-10

.LHS /4.=.RHS /4.

4a/4=-10/4

Calculate quotient

a=-10/4

a/b=.a /2./.b /2.

a=-5/2

a/b=a * 5/b * 5

a=-25/10

Write as a decimal

a = - 2.5

The solution of the equation is a=- 2.5.

Subchapter links

Lesson Check p.206

Practice and Problem-Solving Exercises p.206-208

Standardized Test Prep p.208

Mixed Review p.208