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Q1(ii):
Solve the following pair of linear equations by the substitution method. (ii) s – t = 3; $\frac{s}{3} + \frac{t}{2} = 6$

Solution :

Given: A pair of linear equations in two variables:

(1) $s - t = 3$

(2) $\frac{s}{3} + \frac{t}{2} = 6$

To Find: The values of variables $s$ and $t$ using the substitution method.

Step 1: Express one variable in terms of the other using Equation (1).

From Equation (1):

$s - t = 3$

Adding $t$ to both sides of the equation [Addition Property of Equality]:

$s = t + 3$ --- (Equation 3)

Step 2: Simplify Equation (2) to eliminate fractions.

Equation (2) is given as:

$\frac{s}{3} + \frac{t}{2} = 6$

To add the fractions, find the Least Common Multiple (LCM) of the denominators 3 and 2, which is 6.

Multiply the entire equation by 6 to clear the denominators:

$6 \cdot \left( \frac{s}{3} \right) + 6 \cdot \left( \frac{t}{2} \right) = 6 \cdot 6$

$2s + 3t = 36$ --- (Equation 4)

Step 3: Substitute Equation (3) into Equation (4).

Substitute $s = t + 3$ into $2s + 3t = 36$:

$2(t + 3) + 3t = 36$

Apply the distributive property $a(b+c) = ab + ac$:

$2t + 6 + 3t = 36$

Step 4: Solve for the variable $t$.

Combine like terms ($2t$ and $3t$):

$5t + 6 = 36$

Subtract 6 from both sides [Subtraction Property of Equality]:

$5t = 36 - 6$

$5t = 30$

Divide both sides by 5 [Division Property of Equality]:

$t = \frac{30}{5}$

$t = 6$

Step 5: Solve for the variable $s$ using Equation (3).

Substitute $t = 6$ into Equation (3):

$s = t + 3$

$s = 6 + 3$

$s = 9$

Verification (Optional but recommended):

Substitute $s=9$ and $t=6$ into Equation (2):

$\frac{9}{3} + \frac{6}{2} = 3 + 3 = 6$. The values satisfy the equation.

Final Answer: $s = 9$ and $t = 6$


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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