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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Mid-Chapter Quiz

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Exercise 6 Page 115

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

b=-16

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. In this case, the variable b is already on the right-hand side of the equation, so we will start by subtracting 11 from both sides.

3=1/2b+11

LHS-11=RHS-11

3-11=1/2b+11-11

Subtract terms

-8=1/2b

LHS * 2=RHS* 2

-8*2=1/2b*2

Multiply

-16=b

Rearrange equation

b=-16

The solution to the equation is b=- 16. We can check our solution by substituting it into the original equation.

3=1/2b+11

b= - 16

3 ? = 1/2( - 16)+11

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Simplify

MoveRightFacToNumOne

1/b* a = a/b

3? = - 16/2+11

Calculate quotient

3? = - 8+11

Add terms

3 = 3

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