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Q1(i):
Solve the following pair of linear equations by the elimination method and the substitution method : (i) x + y = 5 and 2x – 3y = 4
Solution :
Given: A pair of linear equations in two variables:
(1) $x + y = 5$
(2) $2x - 3y = 4$
To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.
Method 1: Substitution Method
Step 1: Express one variable in terms of the other.
From equation (1):
$x + y = 5$
$x = 5 - y$ --- (Equation 3)
Step 2: Substitute the expression into the second equation.
Substitute $x = 5 - y$ into equation (2):
$2(5 - y) - 3y = 4$ [Substituting $x$ from Eq 3 into Eq 2]
Step 3: Solve for $y$.
$10 - 2y - 3y = 4$ [Distributive property]
$10 - 5y = 4$ [Combining like terms]
$-5y = 4 - 10$ [Transposing 10 to the RHS]
$-5y = -6$
$y = \frac{-6}{-5}$
$y = \frac{6}{5}$
Step 4: Solve for $x$.
Substitute $y = \frac{6}{5}$ into equation (3):
$x = 5 - \frac{6}{5}$
$x = \frac{25 - 6}{5}$ [Finding a common denominator]
$x = \frac{19}{5}$
Method 2: Elimination Method
Step 1: Make the coefficients of one variable equal.
To eliminate $y$, multiply equation (1) by $3$ so that the coefficient of $y$ in both equations is numerically equal:
$3(x + y) = 3(5)$
$3x + 3y = 15$ --- (Equation 4)
Step 2: Add the equations to eliminate the variable.
Add equation (4) and equation (2):
$(3x + 3y) + (2x - 3y) = 15 + 4$
$3x + 2x + 3y - 3y = 19$ [Grouping like terms]
$5x = 19$
$x = \frac{19}{5}$
Step 3: Substitute the value of $x$ to find $y$.
Substitute $x = \frac{19}{5}$ into equation (1):
$\frac{19}{5} + y = 5$
$y = 5 - \frac{19}{5}$
$y = \frac{25 - 19}{5}$
$y = \frac{6}{5}$
Final Answer: The solution to the system of equations is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.
More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.3
- Q1(ii): Solve the following pair of linear equations by the elimination method and the substitution method : (ii) 3x + 4y = 10 and 2x – 2y = 2
- Q1(iii): Solve the following pair of linear equations by the elimination method and the substitution method : (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
- Q1(iv): Solve the following pair of linear equations by the elimination method and the substitution method : (iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$
- Q2(i): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?
- Q2(ii): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
- Q2(iii): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
- Q2(iv): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her ` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of ` 50 and ` 100 she received.
- Q2(v): Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
CBSE Solutions for Class 10 Mathematics Pair of linear equations in two variable
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