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 $y = 3x - 3$.

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 $t = 6$

Given: A pair of linear equations in two variables:

(1) $0.2x + 0.3y = 1.3$

(2) $0.4x + 0.5y = 2.3$

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

Step 1: Simplifying the equations

To make the calculations easier, we multiply both equations by $10$ to eliminate the decimals.

Equation (1) becomes: $10(0.2x + 0.3y) = 10(1.3) \implies 2x + 3y = 13$ --- (Equation 3)

Equation (2) becomes: $10(0.4x + 0.5y) = 10(2.3) \implies 4x + 5y = 23$ --- (Equation 4)

Step 2: Expressing one variable in terms of the other

From Equation (3), we isolate $x$:

$2x = 13 - 3y$

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

Step 3: Substituting the expression into the second equation

Substitute the value of $x$ from Equation (5) into Equation (4):

$4\left(\frac{13 - 3y}{2}\right) + 5y = 23$

[Since $4/2 = 2$, we simplify the expression]

$2(13 - 3y) + 5y = 23$

Step 4: Solving for $y$

Distribute the $2$ across the terms in the parentheses:

$26 - 6y + 5y = 23$

[Combine like terms: $-6y + 5y = -y$]

$26 - y = 23$

Subtract $26$ from both sides:

$-y = 23 - 26$

$-y = -3$

$y = 3$

Step 5: Solving for $x$

Substitute the value $y = 3$ back into Equation (5):

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

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

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

$x = 2$

Step 6: Verification (Optional but recommended)

Substitute $x=2$ and $y=3$ into the original Equation (1):

$0.2(2) + 0.3(3) = 0.4 + 0.9 = 1.3$ (Correct)

Substitute $x=2$ and $y=3$ into the original Equation (2):

$0.4(2) + 0.5(3) = 0.8 + 1.5 = 2.3$ (Correct)

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

Given:

1. Two angles are supplementary, meaning their sum is $180^\circ$.

2. The larger angle exceeds the smaller angle by $18^\circ$.

To Find:

The measures of the two angles.

Step 1: Defining Variables

Let the larger angle be $x$ degrees.

Let the smaller angle be $y$ degrees.

Step 2: Formulating the Equations

Based on the property of supplementary angles:

$x + y = 180$ --- (Equation 1)

Based on the condition that the larger angle exceeds the smaller by $18^\circ$:

$x = y + 18$ --- (Equation 2)

Step 3: Solving by Substitution Method

We already have $x$ expressed in terms of $y$ in Equation 2. We substitute this expression for $x$ into Equation 1.

Substitute $(y + 18)$ for $x$ in Equation 1:

$(y + 18) + y = 180$

[Combining like terms $y$ and $y$]:

$2y + 18 = 180$

[Subtracting 18 from both sides of the equation]:

$2y = 180 - 18$

$2y = 162$

[Dividing both sides by 2]:

$y = \frac{162}{2}$

$y = 81$

Step 4: Finding the value of $x$

Now, substitute the value of $y = 81$ back into Equation 2:

$x = y + 18$

$x = 81 + 18$

$x = 99$

Step 5: Verification

Check if the sum is $180^\circ$: $99^\circ + 81^\circ = 180^\circ$. (Correct)

Check if the difference is $18^\circ$: $99^\circ - 81^\circ = 18^\circ$. (Correct)

Final Answer: The larger angle is $99^\circ$ and the smaller angle is $81^\circ$.

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 $y = 0$.

Given:

1. The difference between two numbers is $26$.

2. One number is three times the other number.

To Find:

The two numbers.

Step 1: Defining the Variables

Let the larger number be $x$.

Let the smaller number be $y$.

Step 2: Formulating the Linear Equations

Based on the given conditions, we can form the following system of linear equations:

Equation (1): $x - y = 26$ [Since the difference between the two numbers is 26]

Equation (2): $x = 3y$ [Since one number is three times the other]

Step 3: Applying the Substitution Method

The substitution method involves substituting the value of one variable from one equation into the other equation.

We already have $x$ expressed in terms of $y$ in Equation (2): $x = 3y$.

Substitute $x = 3y$ into Equation (1):

$(3y) - y = 26$

Step 4: Solving for $y$

Simplify the expression:

$2y = 26$

Divide both sides by $2$:

$y = \frac{26}{2}$

$y = 13$

Step 5: Solving for $x$

Now, substitute the value of $y = 13$ back into Equation (2) to find $x$:

$x = 3y$

$x = 3(13)$

$x = 39$

Step 6: Verification

Check the results against the original conditions:

Difference: $39 - 13 = 26$ (Condition 1 satisfied)

Ratio: $39 = 3 \times 13$ (Condition 2 satisfied)

Final Answer: The two numbers are 39 and 13.

Given: A system of two linear equations in two variables:

(i) $2x + 3y = 11$

(ii) $2x - 4y = -24$

Additionally, a linear relation: $y = mx + 3$.

To Find: The values of $x$ and $y$ that satisfy the system of equations, and subsequently, the value of the constant $m$.

Step 1: Solving the system of linear equations using the Elimination Method.

We have the system:

$2x + 3y = 11$ --- (Equation 1)

$2x - 4y = -24$ --- (Equation 2)

To eliminate the variable $x$, we subtract Equation 2 from Equation 1:

$(2x + 3y) - (2x - 4y) = 11 - (-24)$

$2x + 3y - 2x + 4y = 11 + 24$ [Distributing the negative sign]

$7y = 35$ [Combining like terms: $2x - 2x = 0$ and $3y + 4y = 7y$]

$y = \frac{35}{7}$ [Dividing both sides by 7]

$y = 5$

Step 2: Finding the value of $x$.

Substitute the value $y = 5$ into Equation 1:

$2x + 3(5) = 11$

$2x + 15 = 11$ [Performing multiplication]

$2x = 11 - 15$ [Transposing 15 to the right side]

$2x = -4$

$x = \frac{-4}{2}$ [Dividing both sides by 2]

$x = -2$

Step 3: Finding the value of $m$.

We are given the equation $y = mx + 3$. Substitute $x = -2$ and $y = 5$ into this equation:

$5 = m(-2) + 3$

$5 = -2m + 3$

$5 - 3 = -2m$ [Transposing 3 to the left side]

$2 = -2m$

$m = \frac{2}{-2}$ [Dividing both sides by -2]

$m = -1$

Final Answer: The solution to the system is $x = -2$ and $y = 5$. The value of $m$ for which $y = mx + 3$ is $m = -1$.

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: $x = 2, y = 3$

Given:

1. The cost of 7 bats and 6 balls is ₹ 3800.

2. The cost of 3 bats and 5 balls is ₹ 1750.

To Find:

The cost of one bat and the cost of one ball.

Step 1: Defining Variables

Let the cost of one bat be $x$ and the cost of one ball be $y$.

Step 2: Formulating the Linear Equations

Based on the given information, we can construct the following system of linear equations:

Equation (1): $7x + 6y = 3800$

Equation (2): $3x + 5y = 1750$

Step 3: Applying the Substitution Method

From Equation (2), express $x$ in terms of $y$:

$3x = 1750 - 5y$

$x = \frac{1750 - 5y}{3}$ --- (Equation 3)

Step 4: Substituting Equation (3) into Equation (1)

Substitute the value of $x$ from Equation (3) into Equation (1):

$7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800$

Multiply the entire equation by 3 to eliminate the denominator:

$7(1750 - 5y) + 18y = 3800 \times 3$

$12250 - 35y + 18y = 11400$

Combine the $y$ terms:

$12250 - 17y = 11400$

Isolate the variable $y$:

$-17y = 11400 - 12250$

$-17y = -850$

$y = \frac{-850}{-17}$

$y = 50$

Step 5: Finding the value of $x$

Substitute $y = 50$ into Equation (3):

$x = \frac{1750 - 5(50)}{3}$

$x = \frac{1750 - 250}{3}$

$x = \frac{1500}{3}$

$x = 500$

Step 6: Verification

Check the values in Equation (1): $7(500) + 6(50) = 3500 + 300 = 3800$. (Correct)

Check the values in Equation (2): $3(500) + 5(50) = 1500 + 250 = 1750$. (Correct)

Final Answer: The cost of one bat is ₹ 500 and the cost of one ball is ₹ 50.

Given:

1. The taxi charges consist of a fixed charge and a charge per kilometer.

2. For a distance of $10\text{ km}$, the total charge is $₹105$.

3. For a distance of $15\text{ km}$, the total charge is $₹155$.

To Find:

1. The fixed charge ($x$).

2. The charge per kilometer ($y$).

3. The total charge for a distance of $25\text{ km}$.

Step 1: Formulating the Linear Equations

Let the fixed charge be $x$ (in ₹) and the charge per kilometer be $y$ (in ₹/km).

Based on the problem statement, the total charge is given by the formula: $\text{Total Charge} = \text{Fixed Charge} + (\text{Distance} \times \text{Charge per km})$.

For the first condition ($10\text{ km}$ for $₹105$):

$x + 10y = 105$ --- (Equation 1)

For the second condition ($15\text{ km}$ for $₹155$):

$x + 15y = 155$ --- (Equation 2)

Step 2: Solving by Substitution Method

From Equation 1, express $x$ in terms of $y$:

$x = 105 - 10y$ --- (Equation 3)

Substitute the value of $x$ from Equation 3 into Equation 2:

$(105 - 10y) + 15y = 155$ [Substituting $x$]

$105 + 5y = 155$ [Combining like terms: $-10y + 15y = 5y$]

$5y = 155 - 105$ [Subtracting 105 from both sides]

$5y = 50$

$y = \frac{50}{5}$ [Dividing both sides by 5]

$y = 10$

Now, substitute $y = 10$ back into Equation 3 to find $x$:

$x = 105 - 10(10)$

$x = 105 - 100$

$x = 5$

Step 3: Calculating the Charge for 25 km

The total charge for $25\text{ km}$ is given by the expression: $x + 25y$.

Substitute $x = 5$ and $y = 10$ into the expression:

$\text{Total Charge} = 5 + 25(10)$

$\text{Total Charge} = 5 + 250$

$\text{Total Charge} = 255$

Final Answer:

The fixed charge is ₹5, the charge per km is ₹10, and the total charge for travelling 25 km is ₹255.

Given:

A fraction where the numerator and denominator are unknown. Let the numerator be $x$ and the denominator be $y$. The fraction is represented as $\frac{x}{y}$.

Condition 1: If 2 is added to both the numerator and the denominator, the fraction becomes $\frac{9}{11}$.

Condition 2: If 3 is added to both the numerator and the denominator, the fraction becomes $\frac{5}{6}$.

To Find:

The value of the fraction $\frac{x}{y}$.

Step 1: Formulating the Linear Equations

Based on Condition 1: $\frac{x + 2}{y + 2} = \frac{9}{11}$

Cross-multiplying to simplify: $11(x + 2) = 9(y + 2)$

$11x + 22 = 9y + 18$

$11x - 9y = 18 - 22$

$11x - 9y = -4$ --- (Equation 1)

Based on Condition 2: $\frac{x + 3}{y + 3} = \frac{5}{6}$

Cross-multiplying to simplify: $6(x + 3) = 5(y + 3)$

$6x + 18 = 5y + 15$

$6x - 5y = 15 - 18$

$6x - 5y = -3$ --- (Equation 2)

Step 2: Solving by Substitution Method

From Equation 2, express $x$ in terms of $y$:

$6x = 5y - 3$

$x = \frac{5y - 3}{6}$ --- (Equation 3)

Step 3: Substituting Equation 3 into Equation 1

$11\left(\frac{5y - 3}{6}\right) - 9y = -4$

Multiply the entire equation by 6 to eliminate the denominator:

$11(5y - 3) - 54y = -24$

$55y - 33 - 54y = -24$

$y - 33 = -24$

$y = -24 + 33$

$y = 9$

Step 4: Finding the value of $x$

Substitute $y = 9$ into Equation 3:

$x = \frac{5(9) - 3}{6}$

$x = \frac{45 - 3}{6}$

$x = \frac{42}{6}$

$x = 7$

Step 5: Conclusion

The numerator $x$ is 7 and the denominator $y$ is 9. Therefore, the fraction is $\frac{7}{9}$.

Final Answer: The required fraction is $\frac{7}{9}$.

Given: A pair of linear equations in two variables:

(1) $x + y = 14$

(2) $x - y = 4$

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

Step 1: Express one variable in terms of the other from one of the equations.

Let us consider equation (1):

$x + y = 14$

Isolating $x$ on the left-hand side [By subtracting $y$ from both sides]:

$x = 14 - y$ --- (Equation 3)

Step 2: Substitute the expression for $x$ into the second equation.

We substitute the value of $x$ from Equation (3) into Equation (2):

$(14 - y) - y = 4$ [Substituting $x = 14 - y$ into $x - y = 4$]

Step 3: Solve the resulting linear equation in one variable ($y$).

Simplify the expression:

$14 - 2y = 4$ [Combining like terms: $-y - y = -2y$]

Subtract 14 from both sides:

$-2y = 4 - 14$

$-2y = -10$

Divide both sides by -2:

$y = \frac{-10}{-2}$

$y = 5$

Step 4: Determine the value of the second variable ($x$).

Substitute the value $y = 5$ back into Equation (3):

$x = 14 - y$

$x = 14 - 5$ [Substituting $y = 5$]

$x = 9$

Step 5: Verification (Optional but recommended).

Check the values in the original equations:

Equation (1): $x + y = 9 + 5 = 14$ (Correct)

Equation (2): $x - y = 9 - 5 = 4$ (Correct)

Final Answer: The solution to the system of equations is $x = 9$ and $y = 5$.

Given:

1. Five years hence (in the future), Jacob's age will be three times that of his son.

2. Five years ago (in the past), Jacob's age was seven times that of his son.

To Find:

The present ages of Jacob and his son.

Step 1: Defining Variables

Let the present age of Jacob be $x$ years.

Let the present age of his son be $y$ years.

Step 2: Formulating the Equations

Case 1: Five years hence

Jacob's age will be $(x + 5)$ and his son's age will be $(y + 5)$.

According to the problem: $(x + 5) = 3(y + 5)$

$x + 5 = 3y + 15$

$x - 3y = 10$ --- (Equation 1)

Case 2: Five years ago

Jacob's age was $(x - 5)$ and his son's age was $(y - 5)$.

According to the problem: $(x - 5) = 7(y - 5)$

$x - 5 = 7y - 35$

$x - 7y = -30$ --- (Equation 2)

Step 3: Solving by Substitution Method

From Equation 1, express $x$ in terms of $y$:

$x = 3y + 10$ --- (Equation 3)

Substitute the value of $x$ from Equation 3 into Equation 2:

$(3y + 10) - 7y = -30$

$3y - 7y + 10 = -30$

$-4y = -30 - 10$

$-4y = -40$

$y = \frac{-40}{-4}$

$y = 10$

Step 4: Finding the value of $x$

Substitute $y = 10$ into Equation 3:

$x = 3(10) + 10$

$x = 30 + 10$

$x = 40$

Step 5: Verification

Five years hence: Jacob will be $45$, son will be $15$. $45 = 3 \times 15$ (Correct).

Five years ago: Jacob was $35$, son was $5$. $35 = 7 \times 5$ (Correct).

Final Answer: The present age of Jacob is 40 years and the present age of his son is 10 years.