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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 $x - \frac{y}{3} = 3$
Solution :
Given: A pair of linear equations in two variables:
(1) $\frac{x}{2} + \frac{2y}{3} = -1$
(2) $x - \frac{y}{3} = 3$
To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.
Part 1: Substitution Method
Step 1: Simplify the given equations.
For equation (1): $\frac{x}{2} + \frac{2y}{3} = -1$. Multiplying the entire equation by the Least Common Multiple (LCM) of 2 and 3, which is 6:
$6 \cdot (\frac{x}{2}) + 6 \cdot (\frac{2y}{3}) = 6 \cdot (-1)$
$3x + 4y = -6$ --- (Equation 3)
For equation (2): $x - \frac{y}{3} = 3$. Multiplying the entire equation by 3:
$3 \cdot (x) - 3 \cdot (\frac{y}{3}) = 3 \cdot (3)$
$3x - y = 9$ --- (Equation 4)
Step 2: Express one variable in terms of the other.
From Equation (4), we can isolate $y$:
$y = 3x - 9$ --- (Equation 5)
Step 3: Substitute Equation (5) into Equation (3).
$3x + 4(3x - 9) = -6$
$3x + 12x - 36 = -6$ [Distributive property]
$15x - 36 = -6$ [Combining like terms]
$15x = 30$ [Adding 36 to both sides]
$x = \frac{30}{15} = 2$
Step 4: Find the value of $y$.
Substitute $x = 2$ into Equation (5):
$y = 3(2) - 9$
$y = 6 - 9 = -3$
Part 2: Elimination Method
Step 1: Align the simplified equations.
(3) $3x + 4y = -6$
(4) $3x - y = 9$
Step 2: Eliminate one variable.
Since the coefficients of $x$ are identical ($3$), we subtract Equation (4) from Equation (3):
$(3x + 4y) - (3x - y) = -6 - 9$
$3x + 4y - 3x + y = -15$ [Distributing the negative sign]
$5y = -15$ [Combining like terms]
$y = \frac{-15}{5} = -3$
Step 3: Solve for the remaining variable.
Substitute $y = -3$ into Equation (4):
$3x - (-3) = 9$
$3x + 3 = 9$
$3x = 9 - 3$
$3x = 6$
$x = \frac{6}{3} = 2$
Final Answer: The solution to the system of equations is $x = 2$ and $y = -3$.
More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.3
- 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
- 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
- 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.
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