Earlier we have seen how minimum of two equations are required for solving linear equation in two variables. Also, we have learnt about the method of eliminations for solving linear equations in two variables.
Under this topic, we’ll learn about a new method “Method Of Substitution”. As the name suggest, in this method of solving linear equation in two variables, we will be solving linear equations by substituting value of any one of the variables from one of the linear equations and then substituting if into the second equation to get an equation in form of only one variable and then solving this linear equation in one variable in one variable by our normal method. Then, the value of this variable is substituted in the first equation to get the value of second variable.
Steps involved in solving linear equations in two variables by method of substitution:
Step I: Examine the question carefully and make sure that two different linear equations are given in same two variables.
Step II: Choose any one of the equation from two given equations and try to find out value of any one variable in terms of another variable.
Step III: Now substitute the value of this variable that we found from first equation into the second equation.
Step IV: As we substitute the value of one variable into the second equation, we’ll find that the equation has been converted into a linear equation in one variable.
Step V: Earlier we have learnt concept of solving linear equation in one variable. Solve the linear equation in one variable hence formed by using the same concept.
Step VI: As we find out the value of one variable, substitute it in the equation of previous variable to find out its value.
In this way, values of variables are calculated using the concept of method of substitution.
To understand the concept in a better way, let us have a look at the examples solved below:
1. Solve for ‘x’ and ‘y’:
x + y = 5.
3x + y = 11.
Solution:
x + y = 5 …………… (i)
3x + y = 11 …………… (ii)
Since we are given two different equations in terms of two different linear equations, let us try to solve them using the concept of method of substitution:
From 1st eq. we find that y = 5 - x.
Substituting value of y in eq. (ii), we get;
3x + 5 - x = 11.
⟹ 2x = 11 - 5
⟹ 2x = 6
⟹ x = 6/2
⟹ x = 3.
Substituting x = 3 in y = 5 – x, we get;
y = 5- x
⟹ y = 5 - 3
⟹ y = 2.
Hence, x = 3 and y = 2.
2. Solve for ‘x’ and ‘y’:
2x + 6y = 8
x - 2y = 15
Solution:
2x + 6y = 8 …………… (i)
x - 2y = 15 …………… (ii)
From the given equations, let us consider first equation and find out value of one of the variables, say ‘x’ from it:
2x = 8 - 6y
⟹ x = 4 - 3y.
Substituting x = 4 - 3y/2 in eq. (ii), we get;
4 - 3y - 2y = 15
⟹ -5y = 15 - 4
⟹ -5y = 11
⟹ y = -11/5.
Substituting y = -11/5 in x = 4 - 3y, we get;
x = 4 - 3(-11/5)
⟹ x = 4 + 33/5
⟹ x = 53/5.
Hence, x = 53/5 and y = -11/5.
3. Solve for ‘x’ and ‘y’:
x + 3y = 10
2x + 3y = 21
Solution:
x + 3y = 10 …………… (i)
2x + 3y = 21 …………… (ii)
Let us look closely at the given equations and we’ll find that both the equations have ‘3y’ in common. So, we will find value of 3y from eq. (i) and substitute it in eq. (ii) and solve for the value of ‘x’. from eq. (i);
x + 3y = 10.
⟹ 3y = 10 - x.
Substituting 3y = 10 - x in eq. (ii), we get;
2x + 3y = 21
⟹ 2x + 10 - x = 21
⟹ x = 21 - 10
⟹ x = 11.
Substituting x = 11 in 3y = 10 – x
3y = 10 – x
⟹ 3y = 10 -11
⟹ 3y = -1
⟹ y = -1/3.
Hence, x = 11 and y = -1/3.
4. Solve for ‘x’ and ‘y’:
5x + y = 20
10x - 2y = 50
Solution:
5x + y = 20 …………… (i)
10x - 2y = 50 …………… (ii)
Let us consider eq. (i) and find out value of ‘y from it and substitute it in eq. (ii). So, from eq. (i);
5x + y =20
⟹ y = 20 – 5x
Substituting y = 20 - 5x in eq. (ii), we get;
10x - 2(20 - 5x) = 50
⟹ 10x - 40 + 10x = 50
⟹ 20x = 50 + 40
⟹ 20x = 90
⟹ x = 90/20
⟹ x = \(\frac{9}{2}\).
Substituting x = \(\frac{9}{2}\) in y = 20 - 5x, we get;
y = 20 – 5(\(\frac{9}{2}\))
⟹ y = 20 - \(\frac{45}{2}\)
⟹ y = \(\frac{40}{2}\) - \(\frac{45}{2}\)
⟹ y = \(\frac{40 - 45}{2}\)
⟹ y = \(\frac{-5}{2}\)
⟹ y = -\(\frac{5}{2}\)
Hence, x = \(\frac{9}{2}\) and y = -\(\frac{5}{2}\)
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