Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically. The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like (
There is every possibility of making mistakes while reading such coordinates.
Is there any alternative method of finding the solution?
As it turns out there is. There are several algebraic methods, which we shall now discuss starting with........
Substitution Method
Look at the below given example to understand the procedure.
Example 4:Solve the following pair of equations by substitution method:
7x – 15y = 2 (1)
x + 2y = 3 (2)
Solution:
Step 1 : We pick either of the equations and write one variable in terms of the other.
Say, we consider the Equation (2), we get:
x + 2y = 3 which can also be written as x =
Step 2 : We now substitute the above value of x in Equation (1) which gives us:
7(3 – 2y) – 15y = 2
21 – 14y – 15y = 2
–
Solving we get, y =
Step 3 : Substituting this value of y in Equation (3) (which we got in Step 1), to find value of x :
x =
Therefore, the solution is x =
Verification: Substitute x =
Now, let's summarise it stepwise:
Step 1 : Find the value of
Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in
Note: Sometimes, you can get statements with
If the statement is false, then the pair of linear equations is
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other
Remark: We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the
Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.
- Let s and t be the ages (in years) of Aftab and his daughter, respectively.
- Thus, s – 7 =
which further simplifies to s – + = 0 (1) - The second equation is: s + 3 =
which further simplifies to s – = (2) - Using eq (2) we get, s = 3t +
- Putting this s value into equation (1), we get: (3t + 6) – 7t + 42 = 0 which gives us t =
. - Putting this value of t in Equation (2), we get: s =
- So, Aftab and his daughter are
and years old, respectively.
In a shop the cost of 2 pencils and 3 erasers is Rs.9 and the cost of 4 pencils and 6 erasers is Rs.18. Find the cost of each pencil and each eraser.
- Let x and y be the cost (in Rs.) of one pencil and one eraser, respectively.
- Thus, the pair of linear equations formed were:
= 9 (1) - The second equation is:
= 18 (2) - Using eq (1) we get, x =
(3) - Putting this x value into equation (2), we get: 18 –
+ 6y = 18. - Thus, this statement is true for all values of y i.e. both the given equations are the same. This means that Equations (1) and (2) have infinitely many solutions.
- We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other?
- Expressing x in terms of y from Equation (1) to get x =
- Substituting this value of x in Equation (2) to get: 2(4 – 2y) + 4y – 12 =
- We know that -4 is not equal to 0. Thus, the given pair of linear equations don't have a solution.
- Thus, the two rails will not cross each other.
Elimination Method
Now let us consider another method of eliminating i.e. removing one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.
Example 8 : The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs.2000 per month, find their monthly incomes.
Solution: Let us denote the incomes of the two person by Rs. 9x and Rs. 7x and their expenditures by Rs. 4y and Rs. 3y respectively. Then the equations formed in the situation is given by :
9x –
Step 1: Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then we get the equations:
27x – 12y = 6000 (3)
28x – 12y = 8000 (4)
Step 2: Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get:
(28x – 27x) – (12y – 12y) = 8000 – 6000
x =
Step 3: Substituting this value of x in (1), we get:
9(2000) – 4y = 2000
y =
So, the solution of the equations is x = 2000, y = 4000.
Therefore, the monthly incomes of the persons are Rs.
Verification: 18000 : 14000 = 9 : 7. Also, the ratio of their expenditures = 18000 – 2000 : 14000 – 2000 = 16000 : 12000 =
👏 Excellent...
Remarks:
- The method used in solving the example above is called the elimination method, because we eliminate one variable first, to get a linear equation in one variable.
In the example above, we eliminated y. We could also have eliminated x. Try doing it that way.
- You could also have used the substitution, or graphical method, to solve this problem. Try doing so, and see which method is more convenient.
Let us now summarise the elimination method :
Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3.
If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e. it is inconsistent.
Step 3: Solve the equation in one variable (x or y) so obtained to get its value.
Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
Now to illustrate it, we shall solve few more examples.
Use elimination method to find all possible solutions of the following pair of linear equations :
2x + 3y = 8 (1)
4x + 6y = 7 (2)
- Let's multiply equation (1) by 2 and equation (2) by 1 to make the coefficients of x equal.
- This makes (1): 4x +
y = (3) while (2) becomes 4x + y = (4) - Subtracting Equation (4) from Equation (3): we get 0 =
- Since it is a false statement, the pair of equations has no solution.
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
- Let the ten’s and the unit’s digits in the first number be x and y, respectively.
- So, the first number may be written as
x + y in the expanded form. - When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number can be denoted by
y + . - Thus, the sum of both the numbers gives us: x +
= (1) - The difference gives us:
– x = (2) - If x – y = 2, then solving (1) and (2) by elimination, we get x =
and y = . In this case, we get the number . - If y – x = 2, then solving (1) and (2) by elimination, we get x =
and y = . In this case, we get the number . - Thus, there are two such numbers 42 and 24.
Verification: Here 42 + 24 =
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