Exercise 61 Page 273

Practice makes perfect

a To solve an equation, using the Properties of Equality 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.

In this case, we need to start by removing the parentheses to simplify the left-hand side of the equation.

6-(3+x)=10

Distr

Distribute -1

6-3-x=10

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

6-3-x=10

SubTerm

Subtract term

3-x=10

SubEqn

LHS-3=RHS-3

- x=7

DivEqn

.LHS /(- 1).=.RHS /(- 1).

x=- 7

The solution to the equation is x=- 7. We can check our solution by substituting it into the original equation.

6-(3+x)=10

Substitute

x= - 7

6-(3+( - 7))? =10

▼

Simplify

AddNeg

a+(- b)=a-b

6-(3-7) ? = 10

SubTerm

Subtract term

6-(- 4)? =10

SubNeg

a-(- b)=a+b

10=10

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

b To solve an equation, using the Properties of Equality 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.

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

100(x+3)=200

Distr

Distribute 100

100x+300=200

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

100x+300=200

SubEqn

LHS-300=RHS-300

100x=- 100

DivEqn

.LHS /100.=.RHS /100.

x=- 1

The solution to the equation is x=- 1. We can check our solution by substituting it into the original equation.

100(x+3)=200

Substitute

x= - 1

100( - 1+3) ? = 200

▼

Simplify

AddTerms

Add terms

100(2) ? = 200

Multiply

200=200

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

c To solve an equation, using the Properties of Equality 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.

In this case, we can start by multiplying the right- and left-hand side of the equation by 3 to remove the fraction.

1/3x+4=x-2

MultEqn

LHS * 3=RHS* 3

x+12=3x-6

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

x+12=3x-6

SubEqn

LHS-12=RHS-12

x=3x-18

SubEqn

LHS-3x=RHS-3x

- 2x = - 18

DivEqn

.LHS /(- 2).=.RHS /(- 2).

x=9

The solution to the equation is x=9. We can check our solution by substituting it into the original equation.

1/3x+4=x-2

Substitute

x= 9

1/3( 9)+4 ? = 9-2

▼

Simplify

MoveRightFacToNum

a/c* b = a* b/c

9/3+4 ? = 9-2

CalcQuot

Calculate quotient

3+4 ? = 9-2

AddSubTerms

Add and subtract terms

7=7

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

d To solve an equation, using the Properties of Equality 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.

We can start by expanding the fraction 45, so it has the same denominator as x+245. Then we will multiply both sides of the equation by 45 to remove the fractions.

4/5=x+2/45

ExpandFrac

a/b=a * 9/b * 9

36/45=x+2/45

MultEqn

LHS * 45=RHS* 45

36=x+2

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

36=x+2

SubEqn

LHS-2=RHS-2

34=x