Exercise 31 Page 547

To solve the proportion, we will start by using the Cross Products Property. Remember that we will need to treat 2x+3 and x-1 as single quantities in the cross multiplication process. From here, we will continue solving for x by using the Distributive Property and the Properties of Equality.

(2x+3)(x-1)=18

Distr

Distribute (2x+3)

x(2x+3)-1(2x+3)=18

▼

Distribute x & - 1

Distr

Distribute x

2x^2+3x-1(2x+3)=18

Distr

Distribute - 1

2x^2+3x-2x-3=18

SubEqn

LHS-18=RHS-18

2x^2+3x-2x-3-18=0

SubTerms

Subtract terms

2x^2+x-21=0

We get a quadratic equation. Now, we can identify the values of a, b, and c. 2x^2+x-21=0 ⇕ 2x^2+ 1x+( - 21)=0 We see that a= 2, b= 1, and c= - 21. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 2)( - 21))/2( 2)

▼

Solve for x

CalcPow

Calculate power

x=- 1±sqrt(1-4(2)(- 21))/2(2)

Multiply

Multiply

x=- 1±sqrt(1-8(- 21))/4

MultNegNegOnePar

- a(- b)=a* b

x=- 1±sqrt(1+168)/4

AddTerms

Add terms

x=- 1±sqrt(169)/4

CalcRoot

Calculate root

x=- 1 ± 13/4

The solutions for this equation are x= - 1 ± 134. Let's separate them into the positive and negative cases.

x=- 1 ± 13/4
x_1=- 1+13/4	x_2=- 1-13/4
x_1=12/4	x_2=- 14/4
x_1=3	x_2=- 3.5

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=3 and x_2=- 3.5.