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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

3. Solving Inequalities Using Multiplication or Division

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Exercise 43 Page 72

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

z=6

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. In this case, we will start by subtracting 12z from both sides.

1/2z+4=5/2z-8

LHS-1/2z=RHS-1/2z

1/2z+4-1/2z=5/2z-8-1/2z

MoveRightFacToNum

a/c* b = a* b/c

1/2z+4-1/2z=5* z/2-8-1* z/2

Multiply

1/2z+4-1/2z=5z/2-8-z/2

Subtract fractions

4=5z-z/2-8

Subtract terms

4=4/2z-8

Calculate quotient

4=2z-8

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

4=2z-8

LHS+8=RHS+8

4+8=2z-8+8

Add terms

12=2z

Rearrange equation

2z=12

.LHS /2.=.RHS /2.

2z/2=12/2

Calculate quotient

z=6

The solution to the equation is z=6. We can check our solution by substituting it into the original equation.

1/2z+4=5/2z-8

z= 6

1/2( 6)+4? =5/2( 6)-8

▼

Evaluate

MoveRightFacToNum

a/c* b = a* b/c

1* 6/2+4? =5* 6/2-8

Multiply

6/2+4? =30/2-8

Calculate quotient

3+4? =15-8

Add and subtract terms

7=7

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

Subchapter links

Communicate Your Answer p.67

Monitoring Progress p.68-70

Exercises p.71-72