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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 15 Page 105

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

n=-4

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.

3n-5=7n+11

LHS-7n=RHS-7n

3n-5-7n=7n+11-7n

Subtract terms

- 4n-5=11

LHS+5=RHS+5

- 4n-5+5=11+5

Add terms

- 4n=16

.LHS /(-4).=.RHS /(-4).

- 4n/- 4=16/-4

Calculate quotient

n=-4

The solution to the equation is n=-4. We can check our solution by substituting it into the original equation.

3n-5=7n+11

n= -4

3( - 4)-5? =7( -4)+11

a(- b)=- a * b

-12-5? =- 28+11

Add and subtract terms

- 17=- 17 ✓

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

Subchapter links

Lesson Check p.105

Practice and Problem-Solving Exercises p.105-108

Standardized Test Prep p.108

Mixed Review p.108