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Q1(iii):
Solve the following pair of linear equations by the substitution method. (iii) 3x – y = 3; 9x – 3y = 9
Solution :
Given: A pair of linear equations in two variables:
(1) $3x - y = 3$
(2) $9x - 3y = 9$
To Find: The solution $(x, y)$ for the given pair of linear equations using the substitution method.
Step 1: Express one variable in terms of the other using Equation (1).
From Equation (1):
$3x - y = 3$
Add $y$ to both sides:
$3x = 3 + y$
Subtract $3$ from both sides:
$y = 3x - 3$ --- (Equation 3)
Step 2: Substitute the expression for $y$ into Equation (2).
Equation (2) is $9x - 3y = 9$.
Substitute $y = 3x - 3$ into Equation (2):
$9x - 3(3x - 3) = 9$
Step 3: Solve the resulting equation.
Distribute the $-3$ across the terms inside the parentheses:
$9x - 9x + 9 = 9$
Combine the $x$ terms ($9x - 9x = 0$):
$0 + 9 = 9$
$9 = 9$
Step 4: Interpret the result.
[Since the variable $x$ has been eliminated and we have arrived at a true statement ($9 = 9$), this indicates that the two equations are dependent and represent the same line.]
Specifically, if we divide Equation (2) by $3$:
$\frac{9x}{3} - \frac{3y}{3} = \frac{9}{3}$
$3x - y = 3$
This is identical to Equation (1). Therefore, the pair of linear equations has infinitely many solutions.
Step 5: General form of the solution.
Since the equations are coincident, any value of $x$ will yield a corresponding value of $y$ that satisfies both equations. We can express the solution set as:
$y = 3x - 3$ for any real number $x$.
Final Answer: The pair of linear equations has infinitely many solutions, represented by the relation $y = 3x - 3$.
More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.2
- Q1(i): Solve the following pair of linear equations by the substitution method. (i) x + y = 14; x – y = 4
- Q1(ii): Solve the following pair of linear equations by the substitution method. (ii) s – t = 3; $\frac{s}{3} + \frac{t}{2} = 6$
- Q1(iv): Solve the following pair of linear equations by the substitution method. (iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
- Q1(v): Solve the following pair of linear equations by the substitution method. (v) $\sqrt{2}x + \sqrt{3}y = 0$; $\sqrt{3}x - \sqrt{8}y = 0$
- Q1(vi): Solve the following pair of linear equations by the substitution method. (vi) $\frac{3x}{2} - \frac{5y}{3} = -2$; $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
- Q2: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘$m$’ for which y = mx + 3.
- Q3(i): Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them.
- Q3(ii): Form the pair of linear equations for the following problems and find their solution by substitution method. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
- Q3(iii): Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball.
- Q3(iv): Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
- Q3(v): Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.
- Q3(vi): Form the pair of linear equations for the following problems and find their solution by substitution method. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
CBSE Solutions for Class 10 Mathematics Pair of linear equations in two variable
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