Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

Step 4 Use the multiplication property of equality to isolate xon the left.


Divide by 4.

Step 5 Check by substituting 4 for xin the original equation.


CHECK Original equation


Let
Simplify.

10 = 10 ✓ True


21 - 12 + 12  10


214 - 52 + 3142  4 + 6 x=4.


21 x- 52 + 3 x=x+ 6


x= 4


4 x


4


=


16


4


SECTION 2.1 Linear Equations in One Variable 51


Eliminate the fractions.
Multiply each side by the LCD, 6.

The solution checks, so the solution set is 5 - 16. NOW TRY


NOW TRY
EXERCISE 3
Solve.


51 x- 42 - 9 = 3 - 21 x+ 162


Alwayscheck
your work.

The solution checks, so is the solution set. NOW TRY


OBJECTIVE 5 Solve linear equations with fractions or decimals.When


fractions or decimals appear as coefficients in equations, our work can be made eas-


ier if we multiply each side of the equation by the least common denominator (LCD)


of all the fractions. This is an application of the multiplication property of equality.


Solving a Linear Equation with Fractions

Solve


Step 1


Step 2 Distributive property


Multiply.
Distributive property
Multiply.
Combine like terms.

Step 3 Add 17.


Combine like terms.

Step 4 Divide by 7.


Step 5 CHECK


Let

Add and subtract in the
numerators.

Simplify each fraction.

- 4 =- 4 ✓ True


1 - 5 - 4


6


6


+


- 10


2


- 4


x=-1.


- 1 + 7


6


+


21 - 12 - 8


2


- 4


x+ 7


6


+


2 x- 8


2


=- 4


x=- 1


7 x


7


=


- 7


7


7 x=- 7


7 x- 17 + 17 =- 24 + 17


7 x- 17 =- 24


x+ 7 + 6 x- 24 =- 24


x+ 7 + 312 x 2 + 31 - 82 =- 24


x+ 7 + 312 x- 82 =- 24


6 a


x+ 7


6


b + 6 a


2 x- 8


2


b = 61 - 42


6 a


x+ 7


6


+


2 x- 8


2


b = 61 - 42


x+ 7

6 +


2 x- 8

2 =-4.


EXAMPLE 4


546


NOW TRY
EXERCISE 4
Solve.


x- 4
4

+


2 x+ 4
8

= 5


NOW TRY ANSWERS



  1. 506 4. 5116

Free download pdf