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.
    If for a data, median is 5 and mode is 4, then mean is equal to :

      • 7
      • 11
      • \(\frac{11}{2}\)
      • \(\frac{14}{3}\)

    • 2.
      If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

        • \( -\frac{2}{3} \)
        • \( \frac{2}{3} \)
        • \( \frac{1}{3} \)
        • \( -\frac{1}{3} \)

      • 3.
        The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

          • 2
          • \(-6\)
          • \(-2\)
          • 6

        • 4.
          Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
          Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

            • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
            • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
            • Assertion (A) is true, but Reason (R) is false.
            • Assertion (A) is false, but Reason (R) is true.

          • 5.
            Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

              • 34
              • 43
              • 38.5
              • 41.5

            • 6.
              Case Study - 1 : A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is 21 cm. Cylindrical pool - Outer diameter is 50 cm and inner diameter is 40 cm. Height of solid base is 14 cm. Height of water filled is 7 cm. Observe the figure and answer the following questions : (i) Determine the total height of the fountain. (ii) Find the volume of the ball. (iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. OR (iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.

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