GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5

Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
1. \( x - 3y - 3 = 0 \), \( 3x - 9y - 2 = 0 \)
2. \( 2x + y = 5 \), \( 3x + 2y = 8 \)
3. \( 3x - 5y = 20 \), \( 6x - 10y = 40 \)
4. \( x - 3y - 7 = 0 \), \( 3x - 3y - 15 = 0 \)
Answer:
1. The given pair of linear equations is:
\( x - 3y - 3 = 0 \) ...(1)
\( 3x - 9y - 2 = 0 \) ...(2)
Here, \( a_1 = 1, b_1 = -3, c_1 = -3 \)
\( a_2 = 3, b_2 = -9, c_2 = -2 \)
We see that \( \frac{a_1}{a_2} = \frac{1}{3} \), \( \frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3} \), and \( \frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2} \)
Therefore, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Hence, the provided pair of linear equations has no solution.
In simple words: When the ratios of 'a' and 'b' terms are equal, but not equal to the ratio of 'c' terms, the two lines are parallel and will never cross. This means there is no solution to the system.

Exam Tip: Remember to express equations in the standard form \(ax + by + c = 0\) before comparing coefficients. Incorrectly setting up the ratios can lead to errors in determining the solution type.

Answer:
2. The given pair of linear equations is:
\( 2x + y = 5 \)
\( 3x + 2y = 8 \)
We can rewrite these equations in standard form:
\( 2x + y - 5 = 0 \)
\( 3x + 2y - 8 = 0 \)
Here, \( a_1 = 2, b_1 = 1, c_1 = -5 \)
\( a_2 = 3, b_2 = 2, c_2 = -8 \)
We observe that \( \frac{a_1}{a_2} = \frac{2}{3} \) and \( \frac{b_1}{b_2} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the given pair of linear equations has a unique solution.
To solve this equation using the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)} \)
\( \implies \frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{4 - 3} \)
\( \implies \frac{x}{2} = \frac{y}{1} = \frac{1}{1} \)
\( \implies x = 2, y = 1 \)
Hence, the necessary solution for the given pair of linear equations is \( x = 2, y = 1 \).
Verification:
Substituting \( x = 2, y = 1 \), we find that both the equations (1) and (2) are satisfied as shown below:
\( 2x + y = 2(2) + 1 = 4 + 1 = 5 \)
\( 3x + 2y = 3(2) + 2(1) = 6 + 2 = 8 \)
The verification shows that our solution is correct.
In simple words: We used a special method to find 'x' and 'y' values that work for both equations. We checked these values by putting them back into the original equations, and they made both equations true.

Exam Tip: Always double-check your cross-multiplication calculations, especially the signs. A small error in sign can lead to a completely different solution. Verification is crucial.

Answer:
3. The given pair of linear equations is:
\( 3x - 5y = 20 \)
\( 6x - 10y = 40 \)
Rewrite these equations in standard form:
\( 3x - 5y - 20 = 0 \) ...(1)
\( 6x - 10y - 40 = 0 \) ...(2)
Here, \( a_1 = 3, b_1 = -5, c_1 = -20 \)
\( a_2 = 6, b_2 = -10, c_2 = -40 \)
We see that \( \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \), and \( \frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2} \)
Thus, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Hence, the given pair of linear equations has infinitely many solutions.
In simple words: When all the ratios of 'a', 'b', and 'c' terms are equal, it means both equations represent the exact same line. So, every point on that line is a solution, giving us endless possibilities.

Exam Tip: Recognizing that two equations are multiples of each other quickly tells you there are infinitely many solutions, saving calculation time. Simplify the ratios carefully.

Answer:
4. The given pair of linear equations is:
\( x - 3y - 7 = 0 \) ...(1)
\( 3x - 3y - 15 = 0 \) ...(2)
Here, \( a_1 = 1, b_1 = -3, c_1 = -7 \)
\( a_2 = 3, b_2 = -3, c_2 = -15 \)
We observe that \( \frac{a_1}{a_2} = \frac{1}{3} \) and \( \frac{b_1}{b_2} = \frac{-3}{-3} = 1 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the given pair of linear equations has a unique solution.
To solve this equation using the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(-3)(-15) - (-3)(-7)} = \frac{y}{(-7)(3) - (-15)(1)} = \frac{1}{(1)(-3) - (3)(-3)} \)
\( \implies \frac{x}{45 - 21} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9} \)
\( \implies \frac{x}{24} = \frac{y}{-6} = \frac{1}{6} \)
\( \implies x = \frac{24}{6} = 4 \)
\( \implies y = \frac{-6}{6} = -1 \)
Hence, the required solution for the given pair of linear equations is \( x = 4, y = -1 \).
Verification:
Substituting \( x = 4, y = -1 \), we find that both equations (1) and (2) are satisfied as shown below:
\( x - 3y - 7 = 4 - 3(-1) - 7 = 4 + 3 - 7 = 0 \)
\( 3x - 3y - 15 = 3(4) - 3(-1) - 15 = 12 + 3 - 15 = 0 \)
The solution is correct.
In simple words: We used the cross-multiplication method to find the specific 'x' and 'y' values that solve both equations. After getting the answer, we checked it by plugging the numbers back into the original problems, and they worked perfectly.

Exam Tip: Be careful with negative signs during multiplication, as they are a common source of error. Write down each step clearly to avoid mistakes.

Question 2.
1. For which values of a and b does the following pair of linear equations have an infinite number of solutions?
\( 2x + 3y = 7 \)
\( (a - b)x + (a + b)y = 3a + b - 2 \)
2. For which value of k will the following pair of linear equations have no solution?
\( 3x + y = 1 \)
\( (2k - 1)x + (k - 1)y = 2k + 1 \)
Answer:
1. The given pair of linear equations is:
\( 2x + 3y = 7 \implies 2x + 3y - 7 = 0 \)
\( (a - b)x + (a + b)y = 3a + b - 2 \implies (a - b)x + (a + b)y - (3a + b - 2) = 0 \)
Here, \( a_1 = 2, b_1 = 3, c_1 = -7 \)
\( a_2 = a - b, b_2 = a + b, c_2 = -(3a + b - 2) \)
Since the given pair of linear equations has infinite solutions, we must have:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \implies \frac{2}{a - b} = \frac{3}{a + b} = \frac{-7}{-(3a + b - 2)} \)
\( \implies \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
From the first two parts:
\( \frac{2}{a - b} = \frac{3}{a + b} \)
\( \implies 2(a + b) = 3(a - b) \)
\( \implies 2a + 2b = 3a - 3b \)
\( \implies a - 5b = 0 \) ......(1)
From the last two parts:
\( \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
\( \implies 3(3a + b - 2) = 7(a + b) \)
\( \implies 9a + 3b - 6 = 7a + 7b \)
\( \implies 2a - 4b - 6 = 0 \)
\( \implies a - 2b - 3 = 0 \) ......(2)
To solve equations (1) and (2) by the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{a}{(-5)(-3) - (-2)(0)} = \frac{b}{(0)(1) - (-3)(1)} = \frac{1}{(1)(-2) - (1)(-5)} \)
\( \implies \frac{a}{15 - 0} = \frac{b}{0 + 3} = \frac{1}{-2 + 5} \)
\( \implies \frac{a}{15} = \frac{b}{3} = \frac{1}{3} \)
Therefore, \( \frac{a}{15} = \frac{1}{3} \implies a = \frac{15}{3} = 5 \)
And \( \frac{b}{3} = \frac{1}{3} \implies b = 1 \)
Hence, the required values of \( a \) and \( b \) are 5 and 1 respectively.
In simple words: For the equations to have endless solutions, the ratios of their 'x', 'y', and constant terms must all be equal. We used this rule to create two new equations for 'a' and 'b', then solved them to find the values that make it true.

Exam Tip: For infinite solutions, remember the condition \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). Set up two separate equality pairs from this triple equality to get two linear equations in 'a' and 'b' and then solve them.

Answer:
2. The given pair of linear equations is:
\( 3x + y = 1 \implies 3x + y - 1 = 0 \)
\( (2k - 1)x + (k - 1)y = 2k + 1 \implies (2k - 1)x + (k - 1)y - (2k + 1) = 0 \)
Here, \( a_1 = 3, b_1 = 1, c_1 = -1 \)
\( a_2 = 2k - 1, b_2 = k - 1, c_2 = -(2k + 1) \)
For having no solution, we must have:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)} \)
\( \implies \frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{1}{2k + 1} \)
From the first two parts, we have:
\( \frac{3}{2k - 1} = \frac{1}{k - 1} \)
\( \implies 3(k - 1) = 1(2k - 1) \)
\( \implies 3k - 3 = 2k - 1 \)
\( \implies 3k - 2k = 3 - 1 \)
\( \implies k = 2 \)
We should also check if \( \frac{1}{k - 1} \neq \frac{1}{2k + 1} \) for \( k = 2 \).
\( \frac{1}{2 - 1} = \frac{1}{1} = 1 \)
\( \frac{1}{2(2) + 1} = \frac{1}{4 + 1} = \frac{1}{5} \)
Since \( 1 \neq \frac{1}{5} \), the condition \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) is satisfied for \( k = 2 \).
Hence, the required value of k is 2.
In simple words: For a system of equations to have no solution, the 'x' and 'y' term ratios must be equal, but this common ratio must not be equal to the ratio of the constant terms. We used the first part of this rule to solve for 'k'.

Exam Tip: For no solution, always verify that the third ratio is indeed unequal, not just that the first two are equal. This confirms the lines are parallel and distinct, not coincident.

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods.
\( 8x + 5y = 9 \)
\( 3x + 2y = 4 \)
Answer:
The given pair of linear equations is:
\( 8x + 5y = 9 \) ...(1)
\( 3x + 2y = 4 \) ...(2)

**I. By Substitution method:**
From equation (2), we can express \( y \) in terms of \( x \):
\( 2y = 4 - 3x \)
\( \implies y = \frac{4 - 3x}{2} \) ......(3)
Substitute this value of \( y \) into equation (1):
\( 8x + 5\left(\frac{4 - 3x}{2}\right) = 9 \)
Multiply the entire equation by 2 to remove the fraction:
\( 16x + 5(4 - 3x) = 18 \)
\( 16x + 20 - 15x = 18 \)
\( x + 20 = 18 \)
\( x = 18 - 20 \)
\( x = -2 \)
Substitute this value of \( x \) into equation (3) to find \( y \):
\( y = \frac{4 - 3(-2)}{2} \)
\( y = \frac{4 + 6}{2} \)
\( y = \frac{10}{2} \)
\( y = 5 \)
So, the solution for the given pair of linear equations is \( x = -2, y = 5 \).
In simple words: In the substitution method, we first rewrite one equation to isolate one variable (like 'y'). Then, we replace that variable in the other equation with its new expression. This helps us solve for one variable, and then we find the second one.

**II. By Cross-multiplication method:**
First, let's write the given pair of linear equations in standard form \( ax + by + c = 0 \):
\( 8x + 5y - 9 = 0 \) ...(3)
\( 3x + 2y - 4 = 0 \) ...(4)
To solve equations (3) and (4) by cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(5)(-4) - (2)(-9)} = \frac{y}{(-9)(3) - (-4)(8)} = \frac{1}{(8)(2) - (3)(5)} \)
\( \implies \frac{x}{-20 + 18} = \frac{y}{-27 + 32} = \frac{1}{16 - 15} \)
\( \implies \frac{x}{-2} = \frac{y}{5} = \frac{1}{1} \)
\( \implies x = -2 \) and \( y = 5 \)
Hence, the required solution for the given pair of linear equations is \( x = -2, y = 5 \).
Verification:
Substituting \( x = -2, y = 5 \), we find that both the equations (1) and (2) are satisfied as shown below:
\( 8x + 5y = 8(-2) + 5(5) = -16 + 25 = 9 \)
\( 3x + 2y = 3(-2) + 2(5) = -6 + 10 = 4 \)
The solution is correct.
In simple words: With cross-multiplication, we organize the coefficients into a special pattern to directly find 'x' and 'y'. Both methods lead to the same answer, proving that our calculations are accurate and consistent.

Exam Tip: Practice both substitution and cross-multiplication methods thoroughly. In exams, you might be asked to use a specific method, so understanding both is key.

Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.
(ii) A fraction becomes \( \frac{1}{3} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Answer:
(i) Let the fixed charge be Rs \( x \) and the cost of food per day be Rs \( y \).
According to the question, we can set up the following equations:
For student A: \( x + 20y = 1000 \) ...(1)
For student B: \( x + 26y = 1180 \) ...(2)
Rewrite these equations in standard form:
\( x + 20y - 1000 = 0 \) ...(3)
\( x + 26y - 1180 = 0 \) ...(4)
To solve these equations by cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(20)(-1180) - (26)(-1000)} = \frac{y}{(-1000)(1) - (-1180)(1)} = \frac{1}{(1)(26) - (1)(20)} \)
\( \implies \frac{x}{-23600 + 26000} = \frac{y}{-1000 + 1180} = \frac{1}{26 - 20} \)
\( \implies \frac{x}{2400} = \frac{y}{180} = \frac{1}{6} \)
Therefore, \( \frac{x}{2400} = \frac{1}{6} \implies x = \frac{2400}{6} = 400 \)
And \( \frac{y}{180} = \frac{1}{6} \implies y = \frac{180}{6} = 30 \)
Hence, the fixed charge is Rs 400 and the cost of food per day is Rs 30.
Verification:
Substituting \( x = 400, y = 30 \), we find that both the equations (1) and (2) are satisfied as shown below:
\( x + 20y = 400 + 20(30) = 400 + 600 = 1000 \)
\( x + 26y = 400 + 26(30) = 400 + 780 = 1180 \)
The solution is correct.
In simple words: We set up equations for fixed and daily food costs based on the information given. Using cross-multiplication, we found the fixed hostel charge to be Rs 400 and the daily food cost to be Rs 30. We then checked these values, and they fit both initial conditions.

Exam Tip: Clearly define your variables (e.g., fixed charge as \(x\), daily cost as \(y\)) at the start. This helps in formulating the equations correctly and prevents confusion.

Answer:
(ii) Let the fraction be \( \frac{x}{y} \).
According to the question, we formulate the equations:
When 1 is subtracted from the numerator, the fraction becomes \( \frac{1}{3} \):
\( \frac{x - 1}{y} = \frac{1}{3} \)
\( \implies 3(x - 1) = y \)
\( \implies 3x - 3 = y \)
\( \implies 3x - y - 3 = 0 \) ...(1)
When 8 is added to its denominator, the fraction becomes \( \frac{1}{4} \):
\( \frac{x}{y + 8} = \frac{1}{4} \)
\( \implies 4x = y + 8 \)
\( \implies 4x - y - 8 = 0 \) ...(2)
To solve equations (1) and (2) by the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(-1)(-8) - (-1)(-3)} = \frac{y}{(-3)(4) - (-8)(3)} = \frac{1}{(3)(-1) - (4)(-1)} \)
\( \implies \frac{x}{8 - 3} = \frac{y}{-12 + 24} = \frac{1}{-3 + 4} \)
\( \implies \frac{x}{5} = \frac{y}{12} = \frac{1}{1} \)
\( \implies x = 5 \) and \( y = 12 \)
Hence, the required fraction is \( \frac{5}{12} \).
Verification:
Substituting \( x = 5, y = 12 \), we find that both the equations (1) and (2) are satisfied as shown below:
For equation 1: \( \frac{x - 1}{y} = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3} \)
For equation 2: \( \frac{x}{y + 8} = \frac{5}{12 + 8} = \frac{5}{20} = \frac{1}{4} \)
The solution is correct.
In simple words: We turned the word problem about a fraction into two linear equations. Using cross-multiplication, we solved these equations to find the numerator and denominator, giving us the fraction \( \frac{5}{12} \). We confirmed our answer by putting it back into the original fraction conditions.

Exam Tip: When setting up equations for fractions, remember to handle operations (subtraction from numerator, addition to denominator) carefully to avoid algebraic errors.

Answer:
(iii) Suppose that Yash gave right answers to \( x \) questions and wrong answers to \( y \) questions. Then, the total number of questions in the test is \( x + y \).
According to the question, we can form two equations:
**Case 1:** Yash scored 40 marks, getting 3 marks for each right answer and losing 1 mark for each wrong answer.
\( 3x - y = 40 \) ......(1)
**Case 2:** If 4 marks were awarded for each correct answer and 2 marks were deducted for each incorrect answer, Yash would have scored 50 marks.
\( 4x - 2y = 50 \) .......(2)
Rewrite these equations in standard form:
\( 3x - y - 40 = 0 \) ......(3)
\( 4x - 2y - 50 = 0 \) ......(4)
To solve equations (3) and (4) by the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(-1)(-50) - (-2)(-40)} = \frac{y}{(-40)(4) - (-50)(3)} = \frac{1}{(3)(-2) - (4)(-1)} \)
\( \implies \frac{x}{50 - 80} = \frac{y}{-160 + 150} = \frac{1}{-6 + 4} \)
\( \implies \frac{x}{-30} = \frac{y}{-10} = \frac{1}{-2} \)
From these equalities, we get:
\( x = \frac{-30}{-2} = 15 \)
And \( y = \frac{-10}{-2} = 5 \)
Hence, Yash gave 15 right answers and 5 wrong answers.
Therefore, the total number of questions in the test = \( x + y = 15 + 5 = 20 \).
Verification:
Substituting \( x = 15, y = 5 \), we find that both the equations (1) and (2) are satisfied as shown below:
\( 3x - y = 3(15) - 5 = 45 - 5 = 40 \)
\( 4x - 2y = 4(15) - 2(5) = 60 - 10 = 50 \)
The solution is correct.
In simple words: We used the given scoring rules to make two equations. By solving these, we found Yash had 15 correct answers and 5 incorrect ones. This means there were 20 questions in total on the test.

Exam Tip: Always be careful with positive and negative signs when forming equations from word problems involving scores, especially for marks deducted for wrong answers.

Answer:
(iv) Let the speeds of the two cars be \( x \) km/hour and \( y \) km/hour respectively.
**Case I: When the cars travel in the same direction.**
Let them meet at point P.
Distance = Speed \( \times \) Time
Distance travelled by the car starting from A in 5 hours = AP = \( 5x \) km
Distance travelled by the car starting from B in 5 hours = BP = \( 5y \) km
Since they start 100 km apart and travel in the same direction, when they meet, the difference in distances must be 100 km.
\( AP - BP = 100 \)
\( \implies 5x - 5y = 100 \)
Dividing by 5:
\( x - y = 20 \) ...(1)
**Case II: When the cars travel towards each other.**
Let they meet at point Q.
Distance travelled by the car starting from A in 1 hour = AQ = \( x \) km
Distance travelled by the car starting from B in 1 hour = BQ = \( y \) km
Since they travel towards each other and start 100 km apart, when they meet, the sum of their distances must be 100 km.
\( AQ + BQ = 100 \)
\( \implies x + y = 100 \) ......(2)
Now, we have a pair of linear equations:
\( x - y = 20 \)
\( x + y = 100 \)
Rewrite these equations in standard form:
\( x - y - 20 = 0 \) ...(3)
\( x + y - 100 = 0 \) ...(4)
To solve the equations (3) and (4) by the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(-1)(-100) - (1)(-20)} = \frac{y}{(-20)(1) - (-100)(1)} = \frac{1}{(1)(1) - (1)(-1)} \)
\( \implies \frac{x}{100 + 20} = \frac{y}{-20 + 100} = \frac{1}{1 + 1} \)
\( \implies \frac{x}{120} = \frac{y}{80} = \frac{1}{2} \)
Therefore, \( x = \frac{120}{2} = 60 \)
And \( y = \frac{80}{2} = 40 \)
Hence, the speeds of the two cars are 60 km/hour and 40 km/hour respectively.
Verification:
Substituting \( x = 60, y = 40 \), we find that both the equations (1) and (2) are satisfied as shown below:
\( x - y = 60 - 40 = 20 \)
\( x + y = 60 + 40 = 100 \)
The solution is correct.
In simple words: We used the meeting times and distances to create two equations for the cars' speeds. When traveling in the same direction, their relative speed gives the first equation; when traveling towards each other, their combined speed gives the second. Solving these equations tells us the individual speeds of each car.

Exam Tip: Remember the basic formula: Distance = Speed × Time. When objects move in the same direction, subtract their speeds; when they move in opposite directions (towards each other), add their speeds.

Answer:
(v) Let the dimensions (i.e., the length and the breadth) of the rectangle be \( x \) units and \( y \) units respectively.
Then, the initial area of the rectangle is \( \text{Length} \times \text{Breadth} = xy \) square units.
According to the question, we form two equations:
**Condition 1:** The area gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units.
New length = \( x - 5 \)
New breadth = \( y + 3 \)
New area = \( (x - 5)(y + 3) \)
So, \( xy - 9 = (x - 5)(y + 3) \)
\( \implies xy - 9 = xy + 3x - 5y - 15 \)
\( \implies 0 - 9 = 3x - 5y - 15 \)
\( \implies 3x - 5y - 6 = 0 \) ...(1)
**Condition 2:** If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units.
New length = \( x + 3 \)
New breadth = \( y + 2 \)
New area = \( (x + 3)(y + 2) \)
So, \( xy + 67 = (x + 3)(y + 2) \)
\( \implies xy + 67 = xy + 2x + 3y + 6 \)
\( \implies 0 + 67 = 2x + 3y + 6 \)
\( \implies 2x + 3y - 61 = 0 \) ...(2)
To solve equations (1) and (2) by the cross-multiplication method, we draw the diagram below:

Then,
\( \frac{x}{(-5)(-61) - (3)(-6)} = \frac{y}{(-6)(2) - (-61)(3)} = \frac{1}{(3)(3) - (2)(-5)} \)
\( \implies \frac{x}{305 + 18} = \frac{y}{-12 + 183} = \frac{1}{9 + 10} \)
\( \implies \frac{x}{323} = \frac{y}{171} = \frac{1}{19} \)
Therefore, \( x = \frac{323}{19} = 17 \)
And \( y = \frac{171}{19} = 9 \)
Hence, the dimensions (i.e., the length and the breadth) of the rectangle are 17 units and 9 units respectively.
Verification:
Substituting \( x = 17, y = 9 \), we find that both the equations (1) and (2) are satisfied as shown below:
For equation (1): \( 3x - 5y - 6 = 3(17) - 5(9) - 6 = 51 - 45 - 6 = 0 \)
For equation (2): \( 2x + 3y - 61 = 2(17) + 3(9) - 61 = 34 + 27 - 61 = 0 \)
The solution is correct.
In simple words: We used the information about changes in the rectangle's dimensions and area to create two linear equations. Solving these equations by cross-multiplication gave us the original length (17 units) and breadth (9 units) of the rectangle.

Exam Tip: Always set up equations from word problems carefully. Pay attention to how changes in length and breadth affect the area, and ensure your algebraic expressions accurately reflect these changes.