1. How to solve algebraic word problems?
A word problem is a math problem that is written in a few sentences that describes a real problem and asks you to apply your math knowledge to solve it.
For example, Laura has 18 coins, she gives 8 of them to John. How many coins are left with Laura?
This is a simple example of a word problem, which can be solved by looking at the numbers and understanding what is asked. If you are able to solve this problem, then you may also be able to solve algebraic word problems. Let's learn the steps to solve an algebra-dependent word problem.
1.1 Identify variables in a word problem
While solving algebraic word problems, the first step is to represent unknown numbers with variables. Some word problems can be solved using only one variable, while for a few complex problems, you might need two variables to solve. You can use a variable in place of any amount you don’t know. Let us now see how to identify the variables with the help of the example below.
“The sum of two consecutive odd numbers is 40. What are the two numbers?”
Since we are trying to work out the two consecutive odd numbers, we’ll represent one of the odd numbers as a variable. If we represent the first odd number as
x, the second consecutive odd number will be x + 2.
1.2 Forming Algebraic Equations
The second prior step in solving algebra dependent word problems is to form algebraic equations using variables by understanding the word problem. After we assign variables to the unknowns in the word problem, we need to create equations in order to solve for these unknown variables and use these unknown values to get the desired answer asked in the word problem. Let us take the same above example and try to create the equation and solve for the unknown variables.
“The sum of two consecutive odd numbers is 40. What are the two numbers?”
We used the variable
x to represent the first odd number and so we have the second odd number as x + 2. The given information in the above example is the sum of these two consecutive odd numbers x and x + 2 is 40.
So we will write equation as x + (x + 2) = 40.
Now we have created the above algebraic equation by assigning the variables to the unknown and by using the information in the problem. You can easily solve the above equation and get the value of x:
x + (x + 2) = 40
x + x = 40 - 2
2x = 38
x = 38 ÷ 2
x = 19
We have assigned the first odd number as
x. So the first odd number is 19. Second odd number is x + 2 which is 19 + 2 = 21. So the two consecutive odd numbers are 19 and 21.
2. Solved Examples
Example:
15 pens and a £2 pack of notebooks cost £5. How much is each pen?
Solution:
As the cost of the pen is unknown, we will represent this with x.
We need 15 pens, so we will represent this with 15x.
The notebooks cost £2 and the total spend is £3.
So our equation is: 15x + 2 = 5
Solving this equation,
15x + 2 = 5
15x = 5 - 2
15x = 3
x = 3 ÷ 15 = 0.2
Therefore the cost of one pen is £0.2 or 10p.
Now let us look at another example of an algebraic word problem which requires two variables.
Example:
5 cupcakes and 4 candy bars cost 100p. 4 cupcakes and 5 candy bars cost 98p. What is the cost of a candy bar and a cupcake?
Solution:
Assume the cost of one cupcake as x and the cost of one candy bar as y.
Cost of 5 cupcakes and 4 candy bars = 5x + 4y = 100p …(1)
Cost of 4 cupcakes and 5 candy bars = 4x + 5y = 98p …(2)
Multiplying (1) by 4, gives 20x + 16y = 400 …(3)
Multiplying (2) by 5, gives 20x + 25y = 490 …(4)
Subtracting equation (4) from (3),
20x + 16y = 400
- 20x + 25y = 490
________________________
-9y = 90
So we get, y = 10
Substitute y = 10 in equation (1),
5x + 4 × 10 = 100
5x + 40 = 100
5x = 100 - 40
5x = 60
x = 12
The cost of one cupcake is 12p and the cost of one candy bar is 10p.
