Pair of Linear Equations in Two Variables | Exercise 3.2

Question 3

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 $\dfrac{9}{11}$ , 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.

(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?

Solution

Understand the Question

To solve real-world problems using a pair of linear equations in two variables:

Identify unknown quantities and assign algebraic variables (e.g., $x$ and $y$ ).
Translate the problem conditions into two linear algebraic equations.
Apply the substitution method: express one variable in terms of the other from one equation, and substitute that expression into the second equation to solve for a single variable.

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

Step 1 · Form Equations and Solve by Substitution

Let the two numbers be $x$ and $y$ (where $x > y$ ).

Given $x - y = 26 \quad \dots (1)$ $x = 3y \quad \dots (2)$

Substitute $(2)$ into $(1)$

\begin{aligned} 3y - y &= 26 \\ 2y &= 26 \\[0.6em] y &= 13 \end{aligned}

Substitute $y = 13$ into $(2)$

x = 3(13) = 39

Answer

(i) $39$ and $13$

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

Step 1 · Form Equations and Solve by Substitution

Let the larger angle be $x$ and the smaller angle be $y$ .

Since supplementary angles add up to $180^\circ$ $x + y = 180 \quad \dots (1)$

Given $x = y + 18 \quad \dots (2)$

Substitute $(2)$ into $(1)$

\begin{aligned} (y + 18) + y &= 180 \\ 2y + 18 &= 180 \\ 2y &= 162 \\[0.6em] y &= 81 \end{aligned}

Substitute $y = 81$ into $(2)$

x = 81 + 18 = 99

Answer

(ii) $99^\circ$ and $81^\circ$

(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.

Step 1 · Form Equations and Express One Variable

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

Given $7x + 6y = 3800 \quad \dots (1)$ $3x + 5y = 1750 \quad \dots (2)$

From $(1)$ $x = \dfrac{3800 - 6y}{7} \quad \dots (3)$

Step 2 · Substitute and Solve for Both Costs

Substitute $(3)$ into $(2)$ $3\left(\dfrac{3800 - 6y}{7}\right) + 5y = 1750$

Multiply by $7$

\begin{aligned} 3(3800 - 6y) + 35y &= 12250 \\ 11400 - 18y + 35y &= 12250 \\ 17y &= 12250 - 11400 \\ 17y &= 850 \\[0.6em] y &= 50 \end{aligned}

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

\begin{aligned} x &= \dfrac{3800 - 6(50)}{7} \\[0.6em] &= \dfrac{3500}{7} \\[0.6em] &= 500 \end{aligned}

Answer

(iii) Cost of a bat $= ₹500$ , cost of a ball $= ₹50$

Step 1 · Find the Fixed Charge and Charge per km

Let the fixed charge be $₹x$ and the charge per $\text{km}$ be $₹y$ .

Given $x + 10y = 105 \quad \dots (1)$ $x + 15y = 155 \quad \dots (2)$

From $(1)$ $x = 105 - 10y \quad \dots (3)$

Substitute $(3)$ into $(2)$

\begin{aligned} (105 - 10y) + 15y &= 155 \\ 5y &= 50 \\[0.6em] y &= 10 \end{aligned}

Substitute $y = 10$ into $(3)$

x = 105 - 10(10) = 5

Step 2 · Calculate Total Charge for 25 km

The charge for travelling $25\text{ km}$ is given by $x + 25y$

\begin{aligned} \text{Charge for } 25\text{ km} &= 5 + 25(10) \\ &= 5 + 250 \\ &= 255 \end{aligned}

Answer

(iv) Fixed charge $= ₹5$ , charge per km $= ₹10$ , charge for $25\text{ km} = ₹255$

Step 1 · Form Equations from the Conditions

Let the fraction be $\dfrac{x}{y}$ .

Condition 1:

\begin{aligned} \dfrac{x+2}{y+2} &= \dfrac{9}{11} \\[0.6em] 11(x+2) &= 9(y+2) \\ 11x + 22 &= 9y + 18 \\ x &= \dfrac{9y - 4}{11} \quad \dots (1) \end{aligned}

Condition 2:

\begin{aligned} \dfrac{x+3}{y+3} &= \dfrac{5}{6} \\[0.6em] 6(x+3) &= 5(y+3) \\ 6x + 18 &= 5y + 15 \\ 6x - 5y &= -3 \quad \dots (2) \end{aligned}

Step 2 · Solve for Numerator and Denominator

Substitute $(1)$ into $(2)$ $6\left(\dfrac{9y - 4}{11}\right) - 5y = -3$

Multiply by $11$

\begin{aligned} 6(9y - 4) - 55y &= -33 \\ 54y - 24 - 55y &= -33 \\ -y &= -33 + 24 \\ y &= 9 \end{aligned}

Substitute $y = 9$ into $(1)$

x = \dfrac{9(9) - 4}{11} = \dfrac{77}{11} = 7

Answer

(v) $\dfrac{7}{9}$

(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?

Step 1 · Form Equations and Solve by Substitution

Let Jacob's present age be $x$ years and his son's present age be $y$ years.

Five years hence:

\begin{aligned} x + 5 &= 3(y + 5) \\ x &= 3y + 10 \quad \dots (1) \end{aligned}

Five years ago:

\begin{aligned} x - 5 &= 7(y - 5) \\ x &= 7y - 30 \quad \dots (2) \end{aligned}

Substitute $(1)$ into $(2)$

\begin{aligned} 3y + 10 &= 7y - 30 \\ 4y &= 40 \\[0.6em] y &= 10 \end{aligned}

Substitute $y = 10$ into $(1)$

x = 3(10) + 10 = 40

Answer

(vi) Jacob's age $= 40\text{ years}$ , son's age $= 10\text{ years}$

Common Mistakes

Supplementary vs. Complementary: Supplementary angles sum to $180^\circ$ , not $90^\circ$ (which is for complementary angles).
Age Shifts: In age word problems, make sure to add or subtract years from both people's ages (e.g., $x+5$ and $y+5$ , not just $x+5$ ).
Brackets in Cross-Multiplication: Forgetting parentheses when multiplying terms like $11(x+2)$ leads to writing $11x + 2$ instead of $11x + 22$ .

Question 3

More questions in Exercise 3.2