Substitution Method of Solving a Pair of Linear Equations

Substitution method is a method of solving Linear Equations. The substitution method is a part of the algebraic method. The substitution method has two classifications that are Algebraic method and the graphical method. 

In the method, we use the value of one variable from an equation and substitute in the other equation. By putting the value, you can easily find the solution. 

EXPLANATION

The substitution method has two classifications that are Algebraic method and the graphical method. 

What is the algebraic method?

Substitution can be done through an algebraic method that means it is a group of different methods by which you can solve the linear equation. The algebraic method includes different variables. We find one variable value and put it in the equation to find the solution. 

How to use substitution methods? 

The equation will be given with two unknown variables. Here are some steps to find the solution and use the substitution method. 

  • First, solve the equation or simplify the equation by expanding them and get the equation in terms of x or y.
  • Now, after getting the equation in terms of x or y. now, solve the equation either for x or for y. 
  • After getting a solution for one variable, now use the substitution method. You have to put the value of the variable in equation one you find in step 1. 
  • Now solve the new equation you get after solving equation 1. 
  • After solving it, you will get the value of the second variable. 
  • After getting both the solution, you can check or verify your answer by putting both the solution in the first equation. 

Difference between substitution method and eliminating method

Here is the difference between the substitution method and eliminating method:

  • In the substitution method, we use the value of one variable from an equation and substitute in the other equation. By putting the value, you can easily find the solution.
  • In eliminating method, we have to add or subtract the equations as eliminating terms to get a value of one variable. After getting a value of one variable, but the value in any equation to get the solution of the second variable. 

A sample example for substitution method

Solve the equations to find ‘x’ and ’y’:

2x + 3y = 21

x + 3y = 10

Answer: 

Put the two equations as 1 & 2.

x + 3y = 10 ……………… (1)

2x + 3y = 21 ……………. (2)

First, solve the equation or simplify the equation by expanding them and get the equation in terms of x or y. solving equation 1

x + 3y = 10

3y = 10 – x ……………… (3)

Now, after getting the equation in terms of x or y. now, solve the equation either for x or for y. Put (3) in equation (2)

2x + 3y = 21

2x + 10 - x = 21

2x – x = 21 – 10

X = 11 

After getting a solution for one variable, now use the substitution method. You have to put the value of the variable in equation one you find in step 1. Put x = 11 in equation 3. 

x + 3y = 10

11 + 3y = 10

3y = 10 – 11

3y = -1

Y = - 1/3

After solving it, you will get the value of the second variable. 

After getting both the solution, you can check or verify your answer by putting both the solution in the first equation. Now, you can put x = 11 and y = -1/3 in equation 1 to verify. 

SAMPLE QUESTIONS

Q1: Solve the equations to find ‘x’ and ’y’:

2x + 3y = 9

x - y = 3

Answer: 

Put the two equations as 1 & 2.

2x + 3y = 9 ……………… (1)

x – y = 3 ……………. (2)

First, solve the equation or simplify the equation by expanding them and get the equation in terms of x or y. solving equation 2

x – y = 3

x = 3 + y ……………… (3)

Now, after getting the equation in terms of x or y. now, solve the equation either for x or for y. Put (3) in equation (1)

2x + 3y = 9

2 (3 + y) + 3y = 9 

6 + 2y + 3y = 9

5y = 9 – 6

5y = 3

Y = 3/5

After getting a solution for one variable, now use the substitution method. You have to put the value of the variable in equation one you find in step 1. Put y = 3/5 in equation 3. 

x = 3 + y 

x = 3+ 3/5

x = 15 + 35

x = 185

After solving it, you will get the value of the second variable. 

After getting both the solution, you can check or verify your answer by putting both the solution in the first equation. Now, you can put x = 185 and y = 3/5 in equation 1 to verify.

Q2: Solve the equations to find ‘x’ and ’y’:

5x + y = 20

10x - 2y = 50

Answer: 

Put the two equations as 1 & 2.

5x + y = 20 ……………… (1)

10x - 2y = 50 ……………. (2)

First, solve the equation or simplify the equation by expanding them and get the equation in terms of x or y. solving equation 1

5x + y = 20

y = 20 – 5x ……………… (3)

Now, after getting the equation in terms of x or y. now, solve the equation either for x or for y. Put (3) in equation (2)

10x - 2y = 50 

10x – 2(20 – 5x) = 50

10x – 40 + 10x = 50

20 x = 50 + 40 

x = 90/ 20

x = 9/2

After getting a solution for one variable, now use the substitution method. You have to put the value of the variable in equation one you find in step 1. Put x = 9/2 in equation 3. 

y = 20 – 5x 

y = 20 - 5 (9/2)

y = 40 - 452

y = -52

After solving it, you will get the value of the second variable. 

After getting both the solution, you can check or verify your answer by putting both the solution in the first equation. Now, you can put x = 9/2 and y = -52 in equation 1 to verify.

CBSE X Related Questions

  • 1.
    Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.


      • 2.
        The value of p for which roots of the quadratic equation $x^2 - px + 6 = 0$ are rational, is

          • $1$
          • $-5$
          • $25$
          • $\sqrt{5}$

        • 3.
          In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.


            • 4.
              Two dice are rolled together. The probability of getting an outcome $(x, y)$ where $x \gt y$, is

                • $\frac{5}{12}$
                • $\frac{5}{6}$
                • $1$
                • $0$

              • 5.
                If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

                  • $x^2 + 5x - 4$
                  • $(x + 3) (-x + 8)$
                  • $a(x^2 + 5x - 24)$
                  • $x^2 - 24$

                • 6.
                  A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

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