RBSE Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.4

Question 1. In a two digit number, the unit digit is three times of tens digit. If 10 is added to two times the number, the digits are reversed. Find the number.
Answer: Let's say the two-digit number is \( 10x + y \), where \( x \) is the digit in the tens place and \( y \) is the digit in the units place.
According to the problem, the unit digit is three times the tens digit:
\( y = 3x \) ...(i)
Also, if 10 is added to two times the number, the digits are reversed. The reversed number would be \( 10y + x \).
So, \( 2(10x + y) + 10 = 10y + x \)
\( 20x + 2y + 10 = 10y + x \)
Now, we rearrange the terms to form a linear equation:
\( 20x - x + 2y - 10y = -10 \)
\( \implies 19x - 8y = -10 \) ...(ii)
Next, substitute the value of \( y \) from equation (i) into equation (ii):
\( 19x - 8(3x) = -10 \)
\( \implies 19x - 24x = -10 \)
\( \implies -5x = -10 \)
Now, divide both sides by -5 to find \( x \):
\( \implies x = \frac{-10}{-5} \)
\( \implies x = 2 \)
Now, substitute the value of \( x \) back into equation (i) to find \( y \):
\( y = 3 \times 2 \)
\( \implies y = 6 \)
So, the tens digit is 2 and the units digit is 6. The number is formed by combining these digits.
Hence, the required number is \( 10x + y = 10(2) + 6 = 20 + 6 = 26 \).
In simple words: We set up equations based on the digits and their relationships. First, we found the tens digit by substituting one equation into another. Then we found the units digit. The number is 26.

🎯 Exam Tip: When dealing with two-digit number problems, always represent the number as \( 10x + y \) and the number with reversed digits as \( 10y + x \), clearly defining \( x \) as the tens digit and \( y \) as the units digit.

Question 2. The perimeter of a rectangle is 56 cm. The ratio of their length and breadth is 4:3. Find the length and breadth of the rectangle.
Answer: Let's denote the length of the rectangle as \( l \) and its breadth as \( b \).
The formula for the perimeter of a rectangle is \( P = 2(l + b) \).
Given that the perimeter is 56 cm, we have:
\( 56 = 2(l + b) \)
Now, divide both sides by 2:
\( \implies l + b = \frac{56}{2} \)
\( \implies l + b = 28 \) ...(i)
We are also given that the ratio of the length and breadth is 4:3. This means we can write:
\( l : b = 4 : 3 \)
This ratio allows us to express \( l \) and \( b \) in terms of a common multiple, say \( x \).
So, let \( l = 4x \) and \( b = 3x \) ...(ii)
Now, substitute these expressions for \( l \) and \( b \) from (ii) into equation (i):
\( 4x + 3x = 28 \)
Combine the terms with \( x \):
\( \implies 7x = 28 \)
Divide by 7 to find the value of \( x \):
\( \implies x = \frac{28}{7} \)
\( \implies x = 4 \)
Now that we have the value of \( x \), we can find the actual length and breadth using the expressions in (ii):
Length \( l = 4x = 4 \times 4 = 16 \) cm
Breadth \( b = 3x = 3 \times 4 = 12 \) cm
Thus, the length of the rectangle is 16 cm and the breadth is 12 cm. This method helps solve problems where measurements are in a specific ratio.
In simple words: We used the perimeter and ratio of length to breadth to find the actual measurements. First, we got an equation from the perimeter. Then, we used the ratio to write length and breadth using 'x'. By putting these into the first equation, we found 'x', and then the length and breadth.

🎯 Exam Tip: Always remember that when a ratio is given (e.g., \( a:b = m:n \)), you can represent the quantities as \( a = mk \) and \( b = nk \) for some constant \( k \). This simplifies setting up equations.

Question 3. Two numbers are in the ratio of 3 : 4. If 5 is subtracted from each number then their ratio become 5: 7. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \).
According to the problem, the ratio of the two numbers is 3:4. This means:
\( x : y = 3 : 4 \)
We can write this as \( \frac{x}{y} = \frac{3}{4} \). To make calculations easier, we can express \( x \) and \( y \) in terms of a common constant, \( k \):
\( x = 3k \) and \( y = 4k \) ...(i)
Next, if 5 is subtracted from each number, their new ratio becomes 5:7. So:
\( \frac{x - 5}{y - 5} = \frac{5}{7} \)
Now, substitute the expressions for \( x \) and \( y \) from (i) into this new ratio equation:
\( \frac{3k - 5}{4k - 5} = \frac{5}{7} \)
To solve for \( k \), we cross-multiply:
\( 7(3k - 5) = 5(4k - 5) \)
\( \implies 21k - 35 = 20k - 25 \)
Now, gather the terms with \( k \) on one side and constant terms on the other:
\( 21k - 20k = -25 + 35 \)
\( \implies k = 10 \)
Finally, substitute the value of \( k \) back into the expressions for \( x \) and \( y \) from (i) to find the numbers:
\( x = 3k = 3 \times 10 = 30 \)
\( y = 4k = 4 \times 10 = 40 \)
So, the two numbers are 30 and 40. This approach works well for ratio-based problems.
In simple words: We used a constant 'k' to represent the numbers based on their first ratio. Then we set up another equation using the second ratio after 5 was subtracted from each number. Solving this equation gave us 'k', which helped us find the actual numbers.

🎯 Exam Tip: For ratio problems, using a common multiple (like \( k \) here) is key. Remember to cross-multiply carefully when solving fractional equations.

Question 4. The present age of a man is five years more than six years more than six times the age of his son. Seven years later, the man's age will be three years more than thrice the age of the sun, then determine their present age.
Answer: Let's assume the present age of the man is \( x \) years and the present age of his son is \( y \) years.
According to the first condition, the man's present age is five years more than six times his son's age:
\( x = 6y + 5 \)
Rearranging this, we get our first equation:
\( \implies x - 6y = 5 \) ...(i)
Now, let's consider their ages seven years later:
Man's age after 7 years: \( x + 7 \)
Son's age after 7 years: \( y + 7 \)
According to the second condition, seven years later, the man's age will be three years more than thrice the age of his son at that time:
\( x + 7 = 3(y + 7) + 3 \)
Distribute the 3 on the right side:
\( \implies x + 7 = 3y + 21 + 3 \)
\( \implies x + 7 = 3y + 24 \)
Rearranging this, we get our second equation:
\( \implies x - 3y = 24 - 7 \)
\( \implies x - 3y = 17 \) ...(ii)
Now we have a system of two linear equations:
1) \( x - 6y = 5 \)
2) \( x - 3y = 17 \)
To solve this system, subtract equation (i) from equation (ii):
\( (x - 3y) - (x - 6y) = 17 - 5 \)
\( \implies x - 3y - x + 6y = 12 \)
\( \implies 3y = 12 \)
Now, divide by 3 to find \( y \):
\( \implies y = \frac{12}{3} \)
\( \implies y = 4 \)
Substitute the value of \( y = 4 \) into equation (i) to find \( x \):
\( x - 6(4) = 5 \)
\( \implies x - 24 = 5 \)
\( \implies x = 5 + 24 \)
\( \implies x = 29 \)
So, the present age of the man is 29 years, and his son's present age is 4 years. Age problems often involve careful tracking of time.
In simple words: We used two equations for the man's and son's ages, one for their current ages and another for their ages seven years later. By solving these two equations together, we found that the man is 29 years old and his son is 4 years old.

🎯 Exam Tip: For age-related word problems, always clearly define variables for present ages first. Then, express ages in the past or future by adding or subtracting years from the present age.

Question 5. Ram said to Shyarn, if you will give Rs. 100 to me them it will be doubles of yours then Shyam said to Ram if you will give me Rs. 10 then my amount will become 6 times. Find what amount they both have?
Answer: Let's assume Ram has Rs. \( x \) and Shyam has Rs. \( y \).
According to Ram's statement: If Shyam gives Rs. 100 to Ram, Ram will have \( x + 100 \), and Shyam will have \( y - 100 \). Ram's amount will be double Shyam's new amount.
So, \( x + 100 = 2(y - 100) \)
\( \implies x + 100 = 2y - 200 \)
Rearranging the terms, we get our first equation:
\( \implies x - 2y = -200 - 100 \)
\( \implies x - 2y = -300 \) ...(i)
According to Shyam's statement: If Ram gives Rs. 10 to Shyam, Ram will have \( x - 10 \), and Shyam will have \( y + 10 \). Shyam's amount will become 6 times Ram's new amount.
So, \( y + 10 = 6(x - 10) \)
\( \implies y + 10 = 6x - 60 \)
Rearranging the terms, we get our second equation:
\( \implies -6x + y = -60 - 10 \)
\( \implies -6x + y = -70 \) ...(ii)
Now we have a system of two linear equations:
1) \( x - 2y = -300 \)
2) \( -6x + y = -70 \)
To solve this, multiply equation (ii) by 2 to make the coefficient of \( y \) the same magnitude as in equation (i):
\( 2(-6x + y) = 2(-70) \)
\( \implies -12x + 2y = -140 \) ...(iii)
Now, add equation (i) and equation (iii):
\( (x - 2y) + (-12x + 2y) = -300 + (-140) \)
\( \implies x - 12x - 2y + 2y = -440 \)
\( \implies -11x = -440 \)
Now, divide by -11 to find \( x \):
\( \implies x = \frac{-440}{-11} \)
\( \implies x = 40 \)
Substitute the value of \( x = 40 \) into equation (i) to find \( y \):
\( 40 - 2y = -300 \)
\( \implies -2y = -300 - 40 \)
\( \implies -2y = -340 \)
Now, divide by -2 to find \( y \):
\( \implies y = \frac{-340}{-2} \)
\( \implies y = 170 \)
Therefore, Ram has Rs. 40 and Shyam has Rs. 170. This problem is a classic example of solving simultaneous equations.
In simple words: We wrote down two equations based on what Ram and Shyam said about exchanging money. We solved these two equations together to find out how much money each person had to start with. Ram had Rs. 40 and Shyam had Rs. 170.

🎯 Exam Tip: Carefully translate the word problem into algebraic equations. Pay attention to "giving" and "receiving" money, as it affects both individuals' amounts in the equations.

Question 6. The cost of 4 chairs and 3 tables is Rs. 2,100 and cost of 5 chairs and 2 tables are Rs. 1,750. Find the cost of 1 chair and 1 table.
Answer: Let's denote the cost of one chair as Rs. \( x \) and the cost of one table as Rs. \( y \).
According to the first statement, the cost of 4 chairs and 3 tables is Rs. 2,100:
\( 4x + 3y = 2100 \) ...(i)
According to the second statement, the cost of 5 chairs and 2 tables is Rs. 1,750:
\( 5x + 2y = 1750 \) ...(ii)
We now have a system of two linear equations. To solve this system by elimination, we can make the coefficients of one variable the same.
Multiply equation (i) by 2:
\( 2 \times (4x + 3y) = 2 \times 2100 \)
\( \implies 8x + 6y = 4200 \) ...(iii)
Multiply equation (ii) by 3:
\( 3 \times (5x + 2y) = 3 \times 1750 \)
\( \implies 15x + 6y = 5250 \) ...(iv)
Now, subtract equation (iii) from equation (iv) to eliminate \( y \):
\( (15x + 6y) - (8x + 6y) = 5250 - 4200 \)
\( \implies 15x - 8x + 6y - 6y = 1050 \)
\( \implies 7x = 1050 \)
Divide by 7 to find the value of \( x \):
\( \implies x = \frac{1050}{7} \)
\( \implies x = 150 \)
Now, substitute the value of \( x = 150 \) into equation (i) to find \( y \):
\( 4(150) + 3y = 2100 \)
\( \implies 600 + 3y = 2100 \)
Subtract 600 from both sides:
\( \implies 3y = 2100 - 600 \)
\( \implies 3y = 1500 \)
Divide by 3 to find the value of \( y \):
\( \implies y = \frac{1500}{3} \)
\( \implies y = 500 \)
Therefore, the cost of one chair is Rs. 150 and the cost of one table is Rs. 500. This is a common method for solving problems with multiple unknown quantities.
In simple words: We made two math sentences (equations) about the cost of chairs and tables. By multiplying the equations and then subtracting them, we could find the cost of one chair. Then we used that number to find the cost of one table.

🎯 Exam Tip: When using the elimination method, always ensure you multiply equations correctly to make the coefficients of one variable equal or opposite, and be careful with signs during subtraction or addition.

Question 7. Two numbers are such that if we divide the 3 times the greater number by the smaller number we obtained the quotient is 4 and remainder 3 and if 7 times the smaller number is divided by bigger number then quotient and remainder are 5 and 1 respectively. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \), and assume \( x \) is the greater number and \( y \) is the smaller number, so \( x > y \).
We use the division algorithm: Dividend = Quotient × Divisor + Remainder.
From the first condition: 3 times the greater number (\( 3x \)) divided by the smaller number (\( y \)) gives a quotient of 4 and a remainder of 3.
So, \( 3x = 4y + 3 \)
Rearranging this, we get our first equation:
\( \implies 3x - 4y = 3 \) ...(i)
From the second condition: 7 times the smaller number (\( 7y \)) divided by the bigger number (\( x \)) gives a quotient of 5 and a remainder of 1.
So, \( 7y = 5x + 1 \)
Rearranging this, we get our second equation:
\( \implies -5x + 7y = 1 \) (or \( 5x - 7y = -1 \)) ...(ii)
Now we have a system of two linear equations:
1) \( 3x - 4y = 3 \)
2) \( -5x + 7y = 1 \)
To solve by elimination, multiply equation (i) by 5 and equation (ii) by 3 to eliminate \( x \):
Multiply (i) by 5: \( 5(3x - 4y) = 5(3) \implies 15x - 20y = 15 \) ...(iii)
Multiply (ii) by 3: \( 3(-5x + 7y) = 3(1) \implies -15x + 21y = 3 \) ...(iv)
Now, add equation (iii) and equation (iv) to eliminate \( x \):
\( (15x - 20y) + (-15x + 21y) = 15 + 3 \)
\( \implies 15x - 15x - 20y + 21y = 18 \)
\( \implies y = 18 \)
Substitute the value of \( y = 18 \) into equation (i) to find \( x \):
\( 3x - 4(18) = 3 \)
\( \implies 3x - 72 = 3 \)
\( \implies 3x = 3 + 72 \)
\( \implies 3x = 75 \)
\( \implies x = \frac{75}{3} \)
\( \implies x = 25 \)
So, the greater number is 25 and the smaller number is 18. This solution relies on using the division algorithm to set up equations.
In simple words: We used the rules of division (dividend, divisor, quotient, remainder) to create two math sentences for the two unknown numbers. By solving these two sentences together, we found that the bigger number is 25 and the smaller number is 18.

🎯 Exam Tip: Clearly identify the dividend, divisor, quotient, and remainder for each division scenario to correctly apply the division algorithm \( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \).

Question 8. A two digit number is 4 times its sum of digits and 2 times the product of its digits. Find the number.
Answer: Let the two-digit number be \( 10x + y \), where \( x \) is the digit in the tens place and \( y \) is the digit in the units place.
According to the first condition, the number is 4 times the sum of its digits:
\( 10x + y = 4(x + y) \)
\( \implies 10x + y = 4x + 4y \)
Rearranging the terms to form an equation:
\( \implies 10x - 4x = 4y - y \)
\( \implies 6x = 3y \)
Dividing both sides by 3, we get:
\( \implies 2x = y \) ...(i)
According to the second condition, the number is 2 times the product of its digits:
\( 10x + y = 2(xy) \) ...(ii)
Now, substitute \( y = 2x \) from equation (i) into equation (ii):
\( 10x + (2x) = 2x(2x) \)
\( \implies 12x = 4x^2 \)
To solve for \( x \), move all terms to one side:
\( \implies 4x^2 - 12x = 0 \)
Factor out \( 4x \):
\( \implies 4x(x - 3) = 0 \)
This gives two possible values for \( x \):
\( 4x = 0 \implies x = 0 \)
or \( x - 3 = 0 \implies x = 3 \)
Since \( x \) is the tens digit of a two-digit number, it cannot be 0 (otherwise, it would be a single-digit number). So, \( x = 3 \).
Now, substitute \( x = 3 \) back into equation (i) to find \( y \):
\( y = 2x = 2(3) = 6 \)
So, the tens digit is 3 and the units digit is 6. This means the number is 36. Verifying the answer shows it satisfies both conditions: \( 36 = 4(3+6) = 4(9) \) and \( 36 = 2(3 \times 6) = 2(18) \).
In simple words: We used two conditions about a two-digit number. First, we wrote the number in terms of its digits. Then, we made two equations based on the sum and product of its digits. By solving these equations, we found that the tens digit is 3 and the units digit is 6, making the number 36.

🎯 Exam Tip: For problems involving digits of a number, remember to represent a two-digit number as \( 10x + y \) and its digits' sum as \( x + y \) and product as \( xy \). Also, remember that the tens digit (\( x \)) cannot be zero for a two-digit number.

Question 9. If 1 is added to both numerator and denominator of a fraction, the fraction becomes \( \frac{4}{5} \). When 5 is subtracted both from numerator and denominator then it become \( \frac{1}{2} \). Find the fraction.
Answer: Let the fraction be \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator.
According to the first condition, if 1 is added to both the numerator and the denominator, the fraction becomes \( \frac{4}{5} \):
\( \frac{x + 1}{y + 1} = \frac{4}{5} \)
Cross-multiplying, we get:
\( 5(x + 1) = 4(y + 1) \)
\( \implies 5x + 5 = 4y + 4 \)
Rearranging the terms, we get our first linear equation:
\( \implies 5x - 4y = 4 - 5 \)
\( \implies 5x - 4y = -1 \) ...(i)
According to the second condition, if 5 is subtracted from both the numerator and the denominator, the fraction becomes \( \frac{1}{2} \):
\( \frac{x - 5}{y - 5} = \frac{1}{2} \)
Cross-multiplying, we get:
\( 2(x - 5) = 1(y - 5) \)
\( \implies 2x - 10 = y - 5 \)
Rearranging the terms, we get our second linear equation:
\( \implies 2x - y = -5 + 10 \)
\( \implies 2x - y = 5 \) ...(ii)
Now we have a system of two linear equations:
1) \( 5x - 4y = -1 \)
2) \( 2x - y = 5 \)
To solve by elimination, multiply equation (ii) by 4 to make the coefficient of \( y \) the same magnitude as in equation (i):
\( 4(2x - y) = 4(5) \)
\( \implies 8x - 4y = 20 \) ...(iii)
Now, subtract equation (i) from equation (iii):
\( (8x - 4y) - (5x - 4y) = 20 - (-1) \)
\( \implies 8x - 5x - 4y + 4y = 20 + 1 \)
\( \implies 3x = 21 \)
Divide by 3 to find the value of \( x \):
\( \implies x = \frac{21}{3} \)
\( \implies x = 7 \)
Substitute the value of \( x = 7 \) into equation (ii) to find \( y \):
\( 2(7) - y = 5 \)
\( \implies 14 - y = 5 \)
\( \implies -y = 5 - 14 \)
\( \implies -y = -9 \)
\( \implies y = 9 \)
Therefore, the numerator is 7 and the denominator is 9. The required fraction is \( \frac{7}{9} \). This problem shows how algebraic equations can represent relationships within fractions.
In simple words: We set up two equations based on how the fraction changes when numbers are added or subtracted from its top and bottom parts. By solving these equations together, we found that the top part of the fraction is 7 and the bottom part is 9, so the fraction is \( \frac{7}{9} \).

🎯 Exam Tip: Always represent the unknown fraction as \( \frac{x}{y} \) and be careful when setting up the equations from word problems involving changes to the numerator and denominator. Cross-multiplication is a crucial step.

Question 10. 5 years ago, Geeta's age was 3 times of Kamla's age. 10 years hence, Geeta's age will be 2 times Kamala's age. Find their present age.
Answer: Let's assume Geeta's present age is \( x \) years and Kamla's present age is \( y \) years.
First condition (5 years ago):
5 years ago, Geeta's age was \( x - 5 \) years.
5 years ago, Kamla's age was \( y - 5 \) years.
According to the problem, 5 years ago, Geeta's age was 3 times Kamla's age:
\( x - 5 = 3(y - 5) \)
\( \implies x - 5 = 3y - 15 \)
Rearranging the terms, we get our first equation:
\( \implies x - 3y = -15 + 5 \)
\( \implies x - 3y = -10 \) ...(i)
Second condition (10 years hence / after 10 years):
10 years from now, Geeta's age will be \( x + 10 \) years.
10 years from now, Kamla's age will be \( y + 10 \) years.
According to the problem, 10 years hence, Geeta's age will be 2 times Kamla's age:
\( x + 10 = 2(y + 10) \)
\( \implies x + 10 = 2y + 20 \)
Rearranging the terms, we get our second equation:
\( \implies x - 2y = 20 - 10 \)
\( \implies x - 2y = 10 \) ...(ii)
Now we have a system of two linear equations:
1) \( x - 3y = -10 \)
2) \( x - 2y = 10 \)
To solve this system, we can subtract equation (i) from equation (ii):
\( (x - 2y) - (x - 3y) = 10 - (-10) \)
\( \implies x - 2y - x + 3y = 10 + 10 \)
\( \implies y = 20 \)
Now, substitute the value of \( y = 20 \) into equation (ii) to find \( x \):
\( x - 2(20) = 10 \)
\( \implies x - 40 = 10 \)
\( \implies x = 10 + 40 \)
\( \implies x = 50 \)
Therefore, Geeta's present age is 50 years and Kamla's present age is 20 years. Age word problems require careful setup of equations based on time frames.
In simple words: We used two equations to figure out Geeta's and Kamla's current ages. One equation was for their ages five years ago, and the other for their ages ten years from now. By solving these, we found Geeta is 50 years old and Kamla is 20 years old.

🎯 Exam Tip: Always establish present ages as variables first. Then, for past events, subtract years; for future events, add years to these variables to form the correct expressions for age at different times.

Question 11. A man travels 370 km partly by train and partly by car. He takes 4 hours if he travels 250 km by train and the remaining by car. If he travels 130 km by train and the remaining by car, he takes 18 minutes longer. Find the speed of the train and the car separately.
Answer: Let the speed of the train be \( x \) km/h and the speed of the car be \( y \) km/h.
The total distance traveled by the man is 370 km. We use the formula: Time = Distance / Speed.
**Case 1:** He travels 250 km by train and the remaining by car.
Distance by train = 250 km
Distance by car = Total distance - Distance by train = \( 370 - 250 = 120 \) km
Time taken by train = \( \frac{250}{x} \) hours
Time taken by car = \( \frac{120}{y} \) hours
Total time taken in this case is 4 hours.
So, \( \frac{250}{x} + \frac{120}{y} = 4 \) ...(i)
**Case 2:** He travels 130 km by train and the remaining by car.
Distance by train = 130 km
Distance by car = Total distance - Distance by train = \( 370 - 130 = 240 \) km
Time taken by train = \( \frac{130}{x} \) hours
Time taken by car = \( \frac{240}{y} \) hours
Total time taken in this case is 18 minutes longer than 4 hours. 18 minutes needs to be converted to hours: \( \frac{18}{60} = \frac{3}{10} \) hours.
So, total time = \( 4 + \frac{18}{60} = 4 + \frac{3}{10} = \frac{40+3}{10} = \frac{43}{10} \) hours.
So, \( \frac{130}{x} + \frac{240}{y} = \frac{43}{10} \) ...(ii)
To solve this system of equations, let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \). The equations become:
1) \( 250a + 120b = 4 \) ...(iii)
2) \( 130a + 240b = \frac{43}{10} \) ...(iv)
Multiply equation (iii) by 2 to make the coefficient of \( b \) equal to 240:
\( 2(250a + 120b) = 2(4) \)
\( \implies 500a + 240b = 8 \) ...(v)
Now, subtract equation (iv) from equation (v) to eliminate \( b \):
\( (500a + 240b) - (130a + 240b) = 8 - \frac{43}{10} \)
\( \implies 500a - 130a + 240b - 240b = \frac{80}{10} - \frac{43}{10} \)
\( \implies 370a = \frac{37}{10} \)
Divide by 370 to find \( a \):
\( \implies a = \frac{37}{10 \times 370} \)
\( \implies a = \frac{1}{100} \)
Now, substitute the value of \( a = \frac{1}{100} \) into equation (iii):
\( 250\left(\frac{1}{100}\right) + 120b = 4 \)
\( \implies \frac{250}{100} + 120b = 4 \)
\( \implies \frac{5}{2} + 120b = 4 \)
Subtract \( \frac{5}{2} \) from both sides:
\( \implies 120b = 4 - \frac{5}{2} \)
\( \implies 120b = \frac{8}{2} - \frac{5}{2} \)
\( \implies 120b = \frac{3}{2} \)
Divide by 120 to find \( b \):
\( \implies b = \frac{3}{2 \times 120} \)
\( \implies b = \frac{3}{240} \)
\( \implies b = \frac{1}{80} \)
Now, substitute back \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \):
\( \frac{1}{x} = \frac{1}{100} \implies x = 100 \)
\( \frac{1}{y} = \frac{1}{80} \implies y = 80 \)
Therefore, the speed of the train is 100 km/h and the speed of the car is 80 km/h. This is a common application of systems of linear equations.
In simple words: We used the formula "time equals distance divided by speed" to set up two equations based on the two different journeys. We introduced new variables to simplify these equations and solved them to find the speeds. The train's speed is 100 km/h and the car's speed is 80 km/h.

🎯 Exam Tip: Always ensure time units are consistent (convert minutes to hours if necessary). When dealing with rational equations, using substitutions like \( a = \frac{1}{x} \) can simplify the system to standard linear equations, making them easier to solve.