124 CHAPTER 2 Linear Equations, Inequalities, and Applications
2.3 Applications of Linear Equations
Solving an Applied Problem
Step 1 Read the problem.
Step 2 Assign a variable.
Step 3 Write an equation.
Step 4 Solve the equation.
Step 5 State the answer.
Step 6 Check.
How many liters of 30% alcohol solution and 80% alcohol solution
must be mixed to obtain 100 L of 50% alcohol solution?
Let
Then 100-x=number of liters of 80% solution needed.
x=number of liters of 30% solution needed.
2.2 Formulas and Percent
Solving a Formula for a Specified Variable (Solving a
Literal Equation)
Step 1 If the equation contains fractions, multiply both
sides by the LCD to clear the fractions.
Step 2 Transform so that all terms with the specified
variable are on one side and all terms without
that variable are on the other side.
Step 3 Divide each side by the factor that is the
coefficient of the specified variable.
Solve for h.
Multiply by 2.
or h= Divide by b.
2 a
b
2 a
b
=h,
2 a=bh
2 a= 2 a
1
2
bhb
a=
1
2
bh
a=
1
2
bh
The equation is
The solution of the equation is 60. Thus, 60 L of 30% solution and
L of 80% solution are needed.
0.30 1602 +0.80 1100 - 602 = 50 is true.
100 - 60 = 40
0.30x+0.80 1100 - x 2 =0.50 11002.
Liters Percent Liters of
of Solution (as a decimal) Pure Alcohol
x 0.30 0.30x
0.80 0.80
100 0.50 0.50 11002
100 - x 1100 - x 2
2.1 Linear Equations in One Variable
Solving a Linear Equation in One Variable
Step 1 Clear fractions.
Step 2 Simplify each side separately.
Step 3 Isolate the variable terms on one side.
Step 4 Isolate the variable.
Step 5 Check.
Solve.
Distributive property
Add 12x.
Subtract 16.
Divide by 4.
The solution set is. This can be checked by substituting 4 for xin
the original equation.
546
4 =x
16
4
=
4 x
4
16 = 4 x
32 - 16 = 16 + 4 x- 16
32 = 16 + 4 x
32 - 12 x+ 12 x= 16 - 8 x+ 12 x
32 - 12 x= 16 - 8 x
32 - 12 x= 32 - 8 x- 16
418 - 3 x 2 = 32 - 81 x+ 22
QUICK REVIEW
CONCEPTS EXAMPLES
(continued)