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Core Connections Algebra 1, 2013

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Core Connections Algebra 1, 2013 View details

2. Section 6.2

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Exercise 74 Page 279

Practice makes perfect

a 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 by using the Properties of Equality.

In this case, we need to start by using the Distributive Property to simplify the left-hand side of the equation.

6(2x-5)=- (x+4)

Distribute 6

12x-30=- (x+4)

Distribute -1

12x-30=- x-4

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

12x-30=- x-4

LHS+30=RHS+30

12x=- x+26

LHS+x=RHS+x

13x=26

.LHS /13.=.RHS /13.

x=2

The solution to the equation is x=2.

b 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 by 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.

(x+1)(x-7)=(x-1)(x+3)

▼

Distribute (x-7)

Distribute (x-7)

x(x-7)+1(x-7)=(x-1)(x+3)

Distribute x

x^2-7x+1(x-7)=(x-1)(x+3)

Distribute 1

x^2-7x+x-7=(x-1)(x+3)

▼

Distribute (x-1)

Distribute (x-1)

x^2-7x+x-7=x(x-1)+3(x-1)

Distribute x

x^2-7x+x-7=x^2-x+3(x-1)

Distribute 3

x^2-7x+x-7=x^2-x+3x-3

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

x^2-7x+x-7=x^2-x+3x-3

LHS-x^2=RHS-x^2

- 7x+x-7=- x+3x-3

Add terms

- 6x-7=2x-3

LHS-2x=RHS-2x

- 8x-7=- 3

LHS+7=RHS+7

- 8x=4

.LHS /(- 8).=.RHS /(- 8).

x=4/- 8

▼

Simplify

MoveNegDenomToFrac

Put minus sign in front of fraction

x=- 4/8

a/b=.a /4./.b /4.

x=- 1/2

The solution to the equation is x=- 12.

Subchapter links

6.2.1 Residual Plots p.271-273

6.2.2 Correlation p.279-280

6.2.3 Association is Not Causation p.283-285

6.2.4 Interpreting Correlation in Context p.290-291

6.2.5 Curved Best-Fit Models p.296-298