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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Two-Step Equations

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Exercise 22 Page 91

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

- 11/2

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.

14=- 2k+3

LHS-3=RHS-3

14-3=- 2k+3-3

Subtract terms

11=- 2k

Rearrange equation

- 2k=11

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

- 2k=11

.LHS /- 2.=.RHS /- 2.

- 2k/- 2=11/- 2

- a/- b=a/b

2k/2=11/- 2

MoveNegDenomToFrac

Put minus sign in front of fraction

2k/2=- 11/2

Calculate quotient

k=- 11/2

The solution to the equation is k=- 112. We can check our solution by substituting it into the original equation.

14=- 2k+3

k= - 11/2

14 ? = - 2( - 11/2)+3

▼

Simplify

MoveLeftFacToNum

a*b/c= a* b/c

14 ? = - - 2(11)/2+3

Multiply

14 ? = - - 22/2+3

MoveNegNumToFrac

Put minus sign in front of fraction

14 ? = - (- 22/2)+3

- (- a)=a

14 ? = 22/2+3

Calculate quotient

14 ? = 11+3

Add terms

14=14

Since the left-hand side is equal to the right-hand side, our solution is correct.

Subchapter links

Lesson Check p.91

Practice and Problem-Solving Exercises p.91-93

Standardized Test Prep p.93

Mixed Review p.93