Solution:

Given Equations:

1) $2x + 3y = 11$

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

Step 1: Solve for x and y using the method of substitution or elimination.

From equation (1): $2x + 3y = 11$

Rearranging for $x$: $2x = 11 - 3y$

$x = \frac{11 - 3y}{2}$

Substitute this value of $x$ into equation (2): $2x - 4y = -24$

Substituting $x = \frac{11 - 3y}{2}$ into the equation:

$2\left(\frac{11 - 3y}{2}\right) - 4y = -24$

Simplifying: $(11 - 3y) - 4y = -24$

$11 - 7y = -24$

$-7y = -24 - 11$

$-7y = -35$

$y = 5$

Step 2: Substitute $y = 5$ into equation (1) to find $x$.

Substitute $y = 5$ into $2x + 3y = 11$:

$2x + 3(5) = 11$

$2x + 15 = 11$

$2x = 11 - 15$

$2x = -4$

$x = -2$

Step 3: Find the value of $m$ for which $y = mx + 3$.

Now, we know that the values of $x = -2$ and $y = 5$. Substitute these into the equation $y = mx + 3$:

$5 = m(-2) + 3$

$5 = -2m + 3$

$-2m = 5 - 3$

$-2m = 2$

$m = -1$

Solution:

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Let the two numbers be $x$ and $y$. The given conditions are: 1. The difference between the two numbers is 26, so $x - y = 26$. 2. One number is three times the other, so $x = 3y$. Substitute $x = 3y$ in the first equation: $3y - y = 26$ $2y = 26$ $y = 13$ Now substitute $y = 13$ in $x = 3y$: $x = 3 \times 13 = 39$ Thus, the two numbers are $39$ and $13$.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the two angles be $x$ and $y$. The given conditions are: 1. The angles are supplementary, so $x + y = 180$. 2. The larger angle exceeds the smaller by 18 degrees, so $x = y + 18$. Substitute $x = y + 18$ in the first equation: $(y + 18) + y = 180$ $2y + 18 = 180$ $2y = 162$ $y = 81$ Now substitute $y = 81$ in $x = y + 18$: $x = 81 + 18 = 99$ Thus, the two angles are $99^\circ$ and $81^\circ$.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

Let the cost of each bat be $x$ and the cost of each ball be $y$. The given conditions are: 1. The cost of 7 bats and 6 balls is Rs. 3800, so $7x + 6y = 3800$. 2. The cost of 3 bats and 5 balls is Rs. 1750, so $3x + 5y = 1750$. Multiply the second equation by 2: $6x + 10y = 3500$ Now subtract the first equation from this: $(6x + 10y) - (7x + 6y) = 3500 - 3800$ $-x + 4y = -300$ $x = 4y + 300$ Substitute $x = 4y + 300$ in $7x + 6y = 3800$: $7(4y + 300) + 6y = 3800$ $28y + 2100 + 6y = 3800$ $34y = 1700$ $y = 50$ Now substitute $y = 50$ in $x = 4y + 300$: $x = 4 \times 50 + 300 = 200 + 300 = 500$ Thus, the cost of each bat is Rs. 500 and the cost of each ball is Rs. 50.

(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 Rs. 105 and for a journey of 15 km, the charge paid is Rs. 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?

Let the fixed charge be $x$ and the charge per km be $y$. The given conditions are: 1. For a distance of 10 km, the total charge is Rs. 105, so $x + 10y = 105$. 2. For a distance of 15 km, the total charge is Rs. 155, so $x + 15y = 155$. $(x + 15y) - (x + 10y) = 155 - 105$ $5y = 50$ $y = 10$ Now substitute $y = 10$ in $x + 10y = 105$: $x + 10 \times 10 = 105$ $x + 100 = 105$ $x = 5$ Thus, the fixed charge is Rs. 5 and the charge per km is Rs. 10. For a distance of 25 km, the total charge will be: $5 + 25 \times 10 = 5 + 250 = 255$ Thus, the person has to pay Rs. 255 for travelling a distance of 25 km.

(v) A fraction becomes $\dfrac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$. Find the fraction.

Let the fraction be $\dfrac{x}{y}$. The given conditions are: 1. If 2 is added to both the numerator and the denominator, the fraction becomes $\dfrac{9}{11}$, so $\dfrac{x + 2}{y + 2} = \dfrac{9}{11}$. 2. If 3 is added to both the numerator and the denominator, the fraction becomes $\dfrac{5}{6}$, so $\dfrac{x + 3}{y + 3} = \dfrac{5}{6}$. From the first equation: $11(x + 2) = 9(y + 2)$ $11x + 22 = 9y + 18$ $11x - 9y = -4$ (Equation 1) From the second equation: $6(x + 3) = 5(y + 3)$ $6x + 18 = 5y + 15$ $6x - 5y = -3$ (Equation 2) Now, solve these two equations: Multiply Equation 1 by 5 and Equation 2 by 9: $55x - 45y = -20$ $54x - 45y = -27$ Subtract the second equation from the first: $55x - 54x = -20 + 27$ $x = 7$ Substitute $x = 7$ in Equation 2: $6(7) - 5y = -3$ $42 - 5y = -3$ $-5y = -45$ $y = 9$ Thus, the fraction is $\dfrac{7}{9}$.

(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?

Let Jacob’s present age be $x$ and his son’s present age be $y$. The given conditions are: 1. Five years hence, Jacob’s age will be three times his son’s age, so $x + 5 = 3(y + 5)$. 2. Five years ago, Jacob’s age was seven times his son’s age, so $x - 5 = 7(y - 5)$. Expanding both equations: From the first equation: $x + 5 = 3y + 15$ $x = 3y + 10$ (Equation 1) From the second equation: $x - 5 = 7y - 35$ $x = 7y - 30$ (Equation 2) Now, solve these two equations: Set Equation 1 equal to Equation 2: $3y + 10 = 7y - 30$ $10 + 30 = 7y - 3y$ $40 = 4y$ $y = 10$ Substitute $y = 10$ in Equation 1: $x = 3(10) + 10 = 30 + 10 = 40$ Thus, Jacob’s present age is 40 years and his son’s present age is 10 years.