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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 26 Page 154

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

Identity.

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 need to start by using the Distributive Property to simplify the left- and right-hand sides of the equation.

3(h-4)=-1/2(24-6h)

Distribute 3

3h-12=-1/2(24-6h)

MoveNegFracToNum

Put minus sign in numerator

3h-12=- 1/2(24-6h)

Distribute - 1/2

3h-12=- 1/2* 24+- 1/2*(- 6h)

MoveRightFacToNum

a/c* b = a* b/c

3h-12=- 24/2+6h/2

Calculate quotient

3h-12=- 12+3h

CommutativePropAdd

Commutative Property of Addition

3h-12=3h-12

Simplifying, the equation resulted in an identity. Therefore, the equation has infinitely many solutions.

Subchapter links

Exercises p.152-156