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Q1(v):
Solve the following pair of linear equations by the substitution method. (v) $\sqrt{2}x + \sqrt{3}y = 0$; $\sqrt{3}x - \sqrt{8}y = 0$
Solution :
Given: A pair of linear equations in two variables:
(i) $\sqrt{2}x + \sqrt{3}y = 0$
(ii) $\sqrt{3}x - \sqrt{8}y = 0$
To find: The values of $x$ and $y$ that satisfy both equations simultaneously using the substitution method.
Step 1: Expressing one variable in terms of the other from Equation (i)
Let the given equations be:
$\sqrt{2}x + \sqrt{3}y = 0$ --- (Equation 1)
$\sqrt{3}x - \sqrt{8}y = 0$ --- (Equation 2)
From Equation (1), isolate $x$:
$\sqrt{2}x = -\sqrt{3}y$
$x = -\frac{\sqrt{3}}{\sqrt{2}}y$ --- (Equation 3)
Step 2: Substituting the expression for $x$ into Equation (2)
Substitute $x = -\frac{\sqrt{3}}{\sqrt{2}}y$ into Equation (2):
$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0$
[Distributing $\sqrt{3}$ into the parenthesis]:
$-\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0$
[Note: $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$]:
$-\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0$
Step 3: Solving for $y$
Factor out $y$ from the expression:
$y \left( -\frac{3}{\sqrt{2}} - 2\sqrt{2} \right) = 0$
[To combine the terms inside the parenthesis, find a common denominator of $\sqrt{2}$]:
$y \left( \frac{-3 - 2\sqrt{2}(\sqrt{2})}{\sqrt{2}} \right) = 0$
$y \left( \frac{-3 - 2(2)}{\sqrt{2}} \right) = 0$
$y \left( \frac{-3 - 4}{\sqrt{2}} \right) = 0$
$y \left( \frac{-7}{\sqrt{2}} \right) = 0$
[Since $\frac{-7}{\sqrt{2}} \neq 0$, we must have $y = 0$]:
$y = 0$
Step 4: Solving for $x$
Substitute the value $y = 0$ back into Equation (3):
$x = -\frac{\sqrt{3}}{\sqrt{2}}(0)$
$x = 0$
Verification:
Substitute $x=0, y=0$ into Equation (1): $\sqrt{2}(0) + \sqrt{3}(0) = 0 + 0 = 0$. (Correct)
Substitute $x=0, y=0$ into Equation (2): $\sqrt{3}(0) - \sqrt{8}(0) = 0 - 0 = 0$. (Correct)
Final Answer: The solution to the system of equations is $x = 0$ and $y = 0$.
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(iii): Solve the following pair of linear equations by the substitution method. (iii) 3x – y = 3; 9x – 3y = 9
- 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(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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