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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Chapter Review

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Exercise 20 Page 153

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

28

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 start by adding 4 to both sides of the equation.

10=c/3-4+c/6

LHS+4=RHS+4

10+4=c/3-4+c/6+4

Add terms

14=c/3+c/6

Now, we can create common denominators for the fractions so that we can add them.

14=c/3+c/6

a/b=a * 2/b * 2

14=2c/6+c/6

Add fractions

14=2c+c/6

Add terms

14=3c/6

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

14=3c/6

LHS * 6=RHS* 6

14*6=3c/6*6

MoveRightFacToNum

a/c* b = a* b/c

14*6=3c*6/6

CancelCommonFac

Cancel out common factors

14*6=3c

Multiply

84=3c

.LHS /3.=.RHS /3.

84/3=3c/3

Calculate quotient

28=c

Rearrange equation

c=28

The solution to the equation is c=28. We can check our solution by substituting it into the original equation.

10=c/3-4+c/6

c= 28

10 ? = 28/3-4+28/6

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

10 ? = 28/3-4+14/3

Add fractions

10 ? = 28+14/3-4

Add terms

10 ? = 42/3-4

Calculate quotient

10 ? = 14-4

Subtract term

10=10

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

Subchapter links

Exercises p.152-156