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Q1(vi):
Solve the following pair of linear equations by the substitution method. (vi) $\frac{3x}{2} - \frac{5y}{3} = -2$; $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Solution :

Given: A pair of linear equations in two variables:

(1) $\frac{3x}{2} - \frac{5y}{3} = -2$

(2) $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

To Find: The values of $x$ and $y$ using the substitution method.

Step 1: Simplifying the equations to standard form.

For equation (1): $\frac{3x}{2} - \frac{5y}{3} = -2$

Find the least common multiple (LCM) of the denominators 2 and 3, which is 6.

Multiply the entire equation by 6: $6 \left( \frac{3x}{2} \right) - 6 \left( \frac{5y}{3} \right) = 6(-2)$

$3(3x) - 2(5y) = -12$

$9x - 10y = -12$ --- (Equation 3)

For equation (2): $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Find the LCM of the denominators 3, 2, and 6, which is 6.

Multiply the entire equation by 6: $6 \left( \frac{x}{3} \right) + 6 \left( \frac{y}{2} \right) = 6 \left( \frac{13}{6} \right)$

$2x + 3y = 13$ --- (Equation 4)

Step 2: Expressing one variable in terms of the other.

From Equation (4), isolate $x$:

$2x = 13 - 3y$

$x = \frac{13 - 3y}{2}$ --- (Equation 5)

Step 3: Substituting Equation (5) into Equation (3).

$9 \left( \frac{13 - 3y}{2} \right) - 10y = -12$

Multiply the entire equation by 2 to eliminate the denominator:

$9(13 - 3y) - 2(10y) = 2(-12)$

$117 - 27y - 20y = -24$ [Using distributive property]

$117 - 47y = -24$

$-47y = -24 - 117$

$-47y = -141$

$y = \frac{-141}{-47}$

$y = 3$

Step 4: Finding the value of $x$.

Substitute $y = 3$ into Equation (5):

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

$x = \frac{4}{2}$

$x = 2$

Verification:

Substitute $x=2, y=3$ into Equation (4): $2(2) + 3(3) = 4 + 9 = 13$. (Correct)

Final Answer: $x = 2, y = 3$


More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.2


CBSE Solutions for Class 10 Mathematics Pair of linear equations in two variable


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