Linear Equations in Word Problems

• Consider the following problem.

The length of the rectangle is 1 unit less than twice its width. If the perimeter of the rectangle is 28 units, what is the width and length of the rectangle? 

To find the width and the length, we can break this process into several steps as follows:

Construct the algebraic expression for “The length of the rectangle is 1 unit less than twice its width ”:

Let \boldsymbol{x} be the width of the rectangle.

Now, twice the width = x × 2 = 2x. Hence, 1 less than twice the width is 2x - 1. Therefore, the length in terms of x is 2x - 1.

Write an algebraic expression for the perimeter using width and length:

Perimeter of rectangle = 2 (length + width)

= 2 \times (2x - 1 + x)

= 2 \times (3x - 1)

= 3x \times 2 -1 \times 2

= 6x - 2

You can also find the perimeter by adding all the sides together. 

Write an equation to represent the problem:

Since we are given the actual perimeter of the rectangle, we can now write an equation by equating the constructed expression for the perimeter to the actual perimeter.

Now, we can solve the equation to find the value of x.

Therefore, the width of the rectangle is 5 units.

We can find the length by substituting the obtained value of x into the 2x - 1 expression.

The length = 2x - 1

      = 2 \times 5 - 1

      = 9

Therefore, the width of the rectangle is \boldsymbol{5} units and the length of the rectangle is \boldsymbol{9} units.

Checking the answer obtained:

Finally, we can check the answer obtained by applying it to the original problem. It was given that the perimeter of the rectangle is 28 units.

Hence,

Since we obtain the same perimeter as given in the problem, we can say that our solution satisfies the original problem. Therefore, our solution is correct.  Similarly, we can construct a linear equation from the word problem and by solving it, we can find the solution to the word problem.

• Let’s consider another example.

John is 5 years younger than his brother. Four years later, John’s brother will be twice as old as John. Find their present ages. 

Let Ron’s present age be x.  As John’s brother is 5 years older, his age in terms of x is x + 5.

After 4 years, John’s age will be x + 4 and his brother’s age will be (x + 5 ) + 4 = x + 9. At that time, John’s brother will be twice as old as John. Therefore, we can multiply John’s age in 4 years by 2 to find John’s brother’s age after 4 years. Thus, John’s brother’s age after 4 years will be 2(x + 4) = 2x + 8. Now, we can construct the following equation considering John’s brother in 4 years:

Now, we solve the equation to find their present ages.

Moreover,

Therefore, John is currently \boldsymbol{1} year old.

John’s brother’s is currently \boldsymbol{5} years old.

• Let’s consider a few more examples.

1. A certain number is 12more than the other. Find the numbers if their sum is 48.

Let x be the smaller number. Thus, the bigger number is x + 12. 

Constructing the equation: 

As the two numbers add to a sum of 48,

Solving the equation: 

Moreover,

∴ Smaller number is \boldsymbol{18}

Bigger number is \boldsymbol{30}

2. Four times a number is 8 less than three times that number. What is the number?

Let x be the number. Four times the number is 4x. Three times the number is 3x and eight less than three times the number is 3x - 8.

Constructing the equation: 

Since the same variable is represented in two different contexts.

Solving the equation: 

∴ The number is \boldsymbol{-8}.

3. The denominator of a fraction is greater than the numerator by 9. If the numerator is increased by 6 and the denominator is decreased by 2, the number obtained is \frac{5}{6}, find the initial fraction.

Let the numerator of the fraction be y. The denominator is 9 more than the numerator; thus, the denominator is y + 9. Increased numerator by 6 is y + 6 and the decreased denominator by 2 is y + 9 - 2 = y + 7.

Constructing the equation: 

After the alterations of the numerator and the denominator, the fraction will be 53:

Solving the equation: 

By cross multiplication,

Swapping sides,

Obtaining the numerator,

Obtaining the denominator,

∴ The initial fraction is \boldsymbol{\frac{-1}{8}}

4. The cost of a sandwich is $0.50 more than the cost of a drink. If the cost of 3 sandwiches and 2 drinks is $6.50, find the cost of each.

Let the cost of the drink be x dollars. The cost of a sandwich is (x + 0.5) dollars. The cost of 3 sandwiches will be 3(x + 0.5) = 3x + 1.5 dollars and the cost of 2 drinks will be 2x dollars.

Constructing the equation: 

The cost of 3 sandwiches and 2 drinks is $6.50:

Solving the equation: 

Moreover,

∴ The cost of a drink is \boldsymbol{$1.0}

 The cost of a sandwich is \boldsymbol{$1.5}

• It is essential to verify the obtained solution for the equation by substituting the calculated value back into the original equation.