Lesson Challenging word problems to solve using a single linear equation

Challenging word problems to solve using a single linear equation

Problem 1. A boat moving upstream and downstream on a river

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 a 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. Mixing water and antifreeze

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 a 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 3. Alloy

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 a 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 4. Web-site

Let x be the number of advertisements that resulted in a sale of the property.


Then 49-x is the number of advertising that did not result in a sale of the property.


Your equation to find x is this revenue equation


    140*x - 30*(49-x) = 3120.


From the equation,


    x =  = 27.    


ANSWER.  The number of advertisements that resulted in a sale of the property is 27.

Problem 5

Let x = # of large boxes;

then the number of small boxes is (120-x).


The total weight equation is

    45x + 35*(120-x) = 4850   pounds.


From the equation

    x =  = 65.


ANSWER.  65 large boxes and 120-65 = 55 small boxes.


CHECK.   45*65 + 35*55 = 4850   (total weight).  ! Precisely correct !