Lesson Using linear equations to solve word problems

Using linear equations to solve word problems

This lesson is a continuation of the lesson How to solve a linear equation, where the general procedure to solve linear equations was described.
In this lesson we present some typical word problems and show how to solve them using linear equations.

Problem 1. Boat moving upstream and downstream on a river

A motorboat moves on a river that has a current of 3 miles per hour.
The trip upstream takes 8 hours, while the return trip takes 5 hours.
What is the speed of the motorboat relative to the water?

Solution
Let us denote the unknown speed of the motorboat relative to the water as x miles/hour.
When motorboat moves upstream, its speed relative to the bank of the river is x-3 miles/hour.
So, moving 8 hours upstream, the motorboat passes 8*(x-3) miles along the river bank.
When moving downstream, the motorboat speed relative to the bank of the river is x+3 miles/hour.
So, moving 5 hours downstream, the motorboat passes 5*(x+3) miles along the river bank.
This is the same distance, so we have an equation

8*(x-3) = 5(x+3).

Note that this is the linear equation.
Performing multiplication and collecting common terms and making other steps of the general procedure, we have

8x - 24 = 5x + 15,
8x - 5x = 24 + 15,
3x = 39,
x = 13 miles/hour.

Check
Substitute 13 to both sides of the original equation. You have:
left side 8*(13-3) = 8*10 = 80,
right side 5*(13+3) = 5*16 = 80.

Answer. The motorboat speed relative to the water is equal to 13 miles/hour.

Problem 2. Alloy

A piece of alloy composed of copper and zinc weights 81400 N (Newtons).
and is 1000 in volume. How much copper and how much zinc is there in the alloy?
The density of copper is 8.92 , the density of zinc is 7.14 .

Solution
Let us denote the unknown volume of copper in the alloy as x .
The weight of copper in the alloy is

981 * 8.92 * x = 8750.52 * x = 8750.52 * x ,

where 981 is the gravity acceleration.
The volume of zinc in alloy is (1000 - x) , and the weight of zinc in alloy is

981 * 7.14 * (1000-x) = 7004.34 * (1000-x) .

The total weight of alloy is 81400 Newtons, which gives us an equation

8750.52 * x + 7004.34 * (1000-x) = 81400 N = 81400 = 8140000 .

(Note that in the previous line we converted Newtons (that are ) to , because we use grams, centimeters and seconds as units uniformly in the solution of this problem).

The last equation is linear one.
Performing multiplication and collecting common terms and making other steps of the general procedure, we have
8750.52 * x + 7004.34*1000 - 7004.34*x = 8140000,
8750.52 * x + 7004340 - 7004.34*x = 8140000,
8750.52 * x - 7004.34*x = 8140000 - 7004340,
1746.18 x = 1135600,
x = 1135600/1746.18 = 650.37
So, the volume of copper is 650.37 , and mass of copper in alloy is 8.92 * 650.37 = 5801 .
Volume of zinc in alloy is 1000-650.37 = 349 , and mass of zinc in alloy is 7.14 * 349 = 2496 .
Answer. Mass of copper in alloy is 5801 g; mass of zinc in alloy is 2496 g.

Problem 3. Mixing water and antifreeze

How much pure antifreeze liquid should be added to 1 gallon of 40% antifreeze to get 60% antifreeze?

(Concentrations here are volume concentrations, measured in [vol/vol] units).

Solution
First, there is 0.4 gal of pure antifreeze in 1 gallon of 40% antifreeze.
Let us denote x a volume of pure antifreeze, which should be added to 1 gallon of 40% antifreeze to get 60% antifreeze.
So, the volume of antifreeze after adding is 0.4 + x gallons, while the total volume of liquid (water plus antifreeze) after adding is 1.0+x.
The condition of 60% volume concentration gives an equation
, or 0.4+x = 0.6*(1.0+x).

This is the linear equation.

Simplifying the equation step by step gives
0.4+x = 0.6 + 0.6*x,
x-0.6*x = 0.6-0.4,
0.4*x = 0.2,
x = 0.2/0.4 = 0.5 gallons.

Check: .
Answer. 0.5 gallons of pure antifreeze should be added.

Problem 4. Mixing hot and cold liquids

There is 400 mL (milliliters, cubical centimeters) of water at 80° C in the thermo-insulated container (thermos).
What volume of water at the temperature 20° C should be added to the container to get the temperature 40° C of the mix water?
Use c = 4.190 for the water specific heat capacity, for the water density.

Solution
Let us denote x this unknown volume of water at the temperature 20° C that should be added to the container to get the temperature 40° C of the mix water. Then the total volume of the liquid in the container is 400 + x .
Based on the energy conservation law, we can write an equation

Substituting all given values, you obtain the equation in the form
1*4.190*(400+x)*40 = 1*4.190*80*x + 1*4.190*400*20.
Reducing by 4.190 both sides and performing other simplifications, you get
(400+x)*40 = 80*x + 400*20,
16000+40x = 80x +8000,
40x-80x = 8000-16000,
-40x = -8000,
x = -8000/(-40) = 200.

Check
(400+200)*40 = 600*40 = 24000 (left side),
80*200 + 400*20 = 16000 + 8000 = 24000 (right side).
The solution is correct.

Answer. 200 milliliters of water at the temperature 20° C should be added.

Note that the density and the specific heat capacity were reduced at the intermediate step and do not produce any influence to the final solution and the final answer.

For more samples of word problems regarding mixtures see the lesson Mixture problems in the section Word problems under the topic Mixtures.