Step 3 Write a system of equations.When the xounces of 5% solution and the
younces of 20% solution are combined, the total number of ounces is 10,
giving the following equation.
(1)The number of ounces of acid in the 5% solution (0.05x) plus the number of
ounces of acid in the 20% solution (0.20y) should equal the total number of
ounces of acid in the mixture, which is (0.125)10, or 1.25.
(2)Notice that these equations can be quickly determined by reading down
the table or using the labels inFIGURE 9.
Step 4 Solvethe system of equations (1) and (2) by eliminating x.
Multiply each side of (1) by.
Multiply each side of (2) by 100.Add.
Divide by 15.Substitute in equation (1) to find that xis also 5.
Step 5 State the answer.The desired mixture will require 5 oz of the 5% solution
and 5 oz of the 20% solution.
Step 6 Check.
Total amount of solution:
as required.
Total amount of acid:
Percent of acid in solution:
or 12.5%, as required.
1.25
10
= 0.125,
=1.25 oz
=0.05 152 +0.20 152
5% of 5 oz+ 20% of 5 oz
= 10 oz,
x+y=5 oz+ 5 oz
y= 5
y= 5
15 y= 75
5 x+ 20 y= 125
- 5 x- 5 y=- 50 -^5
0.05x+ 0.20y=1.25
x+y= 10
SECTION 4.3 Applications of Systems of Linear Equations 237
NOW TRY
EXERCISE 3
How many liters each of a
15% acid solution and a
25% acid solution should be
mixed to get 30 L of an
18% acid solution?
Total acid
Total solution
NOW TRYOBJECTIVE 4 Solve distance-rate-time problems by using two variables.
Motion problems require the distance formula where dis distance, ris rate
(or speed), and tis time.
Solving a Motion ProblemA car travels 250 km in the same time that a truck travels 225 km. If the rate of the car
is 8 km per hr faster than the rate of the truck, find both rates.
Step 1 Readthe problem again. Given the distances traveled, we need to find the
rate of each vehicle.
Step 2 Assign variables.
Let the rate of the car,
andy= the rate of the truck.
x=
EXAMPLE 4
drt,
NOW TRY ANSWER
- 9 L of the 25% acid solution;
21 L of the 15% acid solution