Exercise 41 Page 432

Practice makes perfect

a To solve an equation we should 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.

3x^2+3x=6+3x^2

SubEqn

LHS-3x^2=RHS-3x^2

3x=6

DivEqn

.LHS /3.=.RHS /3.

x=2

The solution to the equation is x=2. We can check our solution by substituting it into the original equation.

3x^2+3x=6+3x^2

Substitute

x= 2

3( 2)^2+3( 2) ? = 6+3( 2)^2

▼

Simplify

CalcPow

Calculate power

3(4)+3(2) ? = 6+3(4)

Multiply

12+6 ? = 6+12

AddTerms

Add terms

18=18 ✓

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

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 using the Properties of Equality. In this case, the variable term is in denominator, so we will start by multiplying the equation by x and then by 3 to eliminate the fractions.

5/x=1/3

MultEqn

LHS * x=RHS* x

5=1/3(x)

MultEqn

LHS * 3=RHS* 3

15=x

RearrangeEqn

Rearrange equation

x=15

The solution to the equation is x=15. We can check our solution by substituting it into the original equation.

5/x=1/3

Substitute

x= 15

5/15 ? = 1/3

ReduceFrac

a/b=.a /5./.b /5.

1/3=1/3 ✓

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

c 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, we first need to remove parentheses to simplify the left-hand side of the equation.

5-(2x-3)=- 3x +6

Distr

Distribute -1

5-2x+3=- 3x+6

AddTerms

Add terms

8-2x=- 3x+6

AddEqn

LHS+3x=RHS+3x

8+x=6

SubEqn

LHS-8=RHS-8

x=- 2

The solution to the equation is x=- 2. We can check our solution by substituting it into the original equation.

5-(2x-3)=- 3x +6

Substitute

x= - 2

5-(2( - 2)-3) ? = - 3( - 2) +6

▼

Simplify

MultPosNeg

a(- b)=- a * b

5-(- 4-3) ? = - 3(- 2) +6

MultNegNegOnePar

- a(- b)=a* b

5-(- 4-3) ? = 6 +6

AddSubTerms

Add and subtract terms

5-(- 7) ? = 12

SubNeg

a-(- b)=a+b

12=12 ✓

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

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

6(x-3)+2x=4(2x+1)-22

Distr

Distribute 6

6x-18+2x=4(2x+1)-22

Distr

Distribute 4

6x-18+2x=8x+4-22

AddSubTerms

Add and subtract terms

8x-18=8x-18

SubEqn

LHS-8x=RHS-8x

- 18 = - 18

Simplifying the equation resulted in an identity. It means that all real numbers satisfy the equation. Thus, the equation has infinitely many solutions.

Subchapter links

9.1.1 Solving Quadratic Equations p.420

9.1.2 Introduction to the Quadratic Formula p.423-424

9.1.3 More Solving Quadratic Equations p.427-428

9.1.4 Choosing a Strategy p.432

Exercise 41 Page 432

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