SECTION 2.7 Further Applications of Linear Equations^141
Step 3 Write an equation.The number of liters of pure acid in the 70% solution
added to the number of liters of pure acid in the 40% solution will equal the
number of liters of pure acid in the final mixture.
Pure acid pure acid pure acid
in 70% plus in 40% is in 50%.0.70x
Step 4 Solvethe equation.
Distributive property
Multiply by 100.
Multiply.
Subtract 50x.
Subtract 800.
Divide by 20.Step 5 State the answer.The chemist needs to use 10 L of 70% solution.
Step 6 Check.The answer checks, since
Sum of two solutionsand
0.50 110 + 202 = 0.50 1302 = 15. Mixture NOW TRY
0.70 1102 +0.40 1202 = 7 + 8 = 15
x= 10
20 x= 200
20 x+ 800 = 1000
70 x+ 800 = 50 x+ 1000
70 x+ 401202 = 50 x+ 501202
0.70x+0.40 1202 =0.50x+ 0.50 1202
+ 0.40 1202 = 0.50 1 x+ 202
NOW TRY
EXERCISE 2
A certain seasoning is 70%
salt. How many ounces of this
seasoning must be mixed with
30 oz of dried herbs contain-
ing 10% salt to obtain a
seasoning that is 50% salt?
Amount Invested Rate of Interest for
in Dollars Interest One Year
x 0.06 0.06x
x+ 2000 0.07 0.07 1 x+ 20002Use a table to arrange the
given information.NOTE In a mixture problem, the concentration of the final mixture must be between
the concentrations of the two solutions making up the mixture.
OBJECTIVE 3 Solve problems involving simple interest.The formula for sim-
ple interest, becomes when time (for annual interest), as shown in
the Problem-Solving Hint at the beginning of this section. Multiplying the total amount
(principal) by the rate (rate of interest) gives the percentage (amount of interest).
I=prt, I= pr t= 1
Solving a Simple Interest ProblemSusan Grody plans to invest some money at 6% and $2000 more than this amount at
7%. To earn $790 per year in interest, how much should she invest at each rate?
Step 1 Readthe problem again. There will be two answers.
Step 2 Assign a variable.
Let the amount invested at 6% (in dollars).
Then x+ 2000 = the amount invested at 7% (in dollars).
x=
EXAMPLE 3
NOW TRY ANSWER
- 60 oz