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. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (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. (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? (v) A fraction becomes 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 5 6 . Find the fraction. 34 (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? 3.3.2 Elimination Method Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.
Step-by-Step Solution
Key Concept: Introduce variables for the unknown quantities, translate the word statements into linear equations, and then solve the resulting pair of equations using the substitution method as described in the NCERT textbook.
### (i) Two numbers
Let the two numbers be $x$ and $y$.
- Difference: $x - y = 26$ … (1)
- One number is three times the other: $x = 3y$ … (2)
Substitution: From (2), $x = 3y$. Substitute in (1):
$$3y - y = 26 \ \ 2y = 26 \ y = 13$$
Then $x = 3y = 39$.\
Answer: The numbers are $39$ and $13$.
### (ii) Supplementary angles
Let the smaller angle be $x^{\circ}$ and the larger be $y^{\circ}$.
- Supplementary: $x + y = 180$ … (1)
- Larger exceeds smaller by $18^{\circ}$: $y = x + 18$ … (2)
Substitution: Put (2) in (1):
$$x + (x + 18) = 180 \ 2x + 18 = 180 \ 2x = 162 \ x = 81$$
Hence $y = 81 + 18 = 99$.\
Answer: Angles are $81^{\circ}$ and $99^{\circ}$.
### (iii) Cost of bats and balls
Let cost of one bat = $b$ rupees, cost of one ball = $c$ rupees.
- First purchase: $7b + 6c = 3800$ … (1)
- Second purchase: $3b + 5c = 1750$ … (2)
Substitution: From (2), express $b$:
$$3b = 1750 - 5c \ b = \frac{1750 - 5c}{3}$$
Substitute in (1):
$$7\left(\frac{1750 - 5c}{3}\right) + 6c = 3800$$
Multiply by 3:
$$7(1750 - 5c) + 18c = 11400$$
$$12250 - 35c + 18c = 11400$$
$$-17c = -850 \ c = 50$$
Then $b = \frac{1750 - 5\times 50}{3} = \frac{1750 - 250}{3} = \frac{1500}{3} = 500$.\
Answer: One bat costs Rs. 500 and one ball costs Rs. 50.
### (iv) Taxi charges
Let fixed charge = $F$ rupees, charge per km = $p$ rupees.
- For 10 km: $F + 10p = 105$ … (1)
- For 15 km: $F + 15p = 155$ … (2)
Substitution: From (1), $F = 105 - 10p$. Substitute in (2):
$$105 - 10p + 15p = 155 \ 105 + 5p = 155 \ 5p = 50 \ p = 10$$
Then $F = 105 - 10\times 10 = 5$.\
Charges: Fixed charge = Rs. 5, charge per km = Rs. 10.
Cost for 25 km: $F + 25p = 5 + 25\times 10 = 5 + 250 = 255$ rupees.
### (v) Fraction problem
Let the required fraction be $\frac{x}{y}$ (with $x, y$ positive integers).
- Adding 2 to numerator and denominator gives $\frac{x+2}{y+2}=\frac{9}{11}$ … (1)
- Adding 3 gives $\frac{x+3}{y+3}=\frac{5}{6}$ … (2)
From (1): Cross‑multiply:
$$11(x+2) = 9(y+2) \ 11x + 22 = 9y + 18 \ 11x - 9y = -4 \ \ (A)$$
From (2):
$$6(x+3) = 5(y+3) \ 6x + 18 = 5y + 15 \ 6x - 5y = -3 \ \ (B)$$
Substitution: Solve (B) for $x$: $6x = 5y - 3 \ x = \frac{5y - 3}{6}$.
Substitute in (A):
$$11\left(\frac{5y - 3}{6}\right) - 9y = -4$$
Multiply by 6:
$$11(5y - 3) - 54y = -24 \ 55y - 33 - 54y = -24 \ y - 33 = -24 \ y = 9$$
Then $x = \frac{5\times 9 - 3}{6} = \frac{45 - 3}{6} = \frac{42}{6} = 7$.
Answer: The fraction is $\frac{7}{9}$.
### (vi) Ages of Jacob and his son
Let Jacob’s present age = $J$ years, son’s present age = $S$ years.
- Five years hence: $J+5 = 3(S+5)$ … (1)
- Five years ago: $J-5 = 7(S-5)$ … (2)
From (1): $J+5 = 3S + 15 \ J = 3S + 10$ … (3)
Substitution: Put (3) in (2):
$$(3S + 10) - 5 = 7(S - 5) \ 3S + 5 = 7S - 35 \ 5 + 35 = 7S - 3S \ 40 = 4S \ S = 10$$
Then $J = 3\times 10 + 10 = 40$.\
Answer: Jacob is 40 years old and his son is 10 years old.
Correct Answer: {"i":{"numbers":[39,13]},"ii":{"angles_degrees":[81,99]},"iii":{"bat_cost_rupees":500,"ball_cost_rupees":50},"iv":{"fixed_charge_rupees":5,"charge_per_km_rupees":10,"cost_for_25km_rupees":255},"v":{"fraction":"7/9"},"vi":{"Jacob_age_years":40,"son_age_years":10}}