Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.11

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.11

Question 1. Solve, using the method of substitution.
(i) \( 2x - 3y = 7 \); \( 5x + y = 9 \)
(ii) \( 1.5x + 0.1y = 6.2 \); \( 3x - 0.4y = 11.2 \)
(iii) \( 10\% \text{ of } x + 20\% \text{ of } y = 24 \); \( 3x - y = 20 \)
(iv) \( \sqrt{2}x - \sqrt{3}y = 1 \); \( \sqrt{3}x - \sqrt{8}y = 0 \)
Answer:
(i) The given equations are:
\( 2x - 3y = 7 \quad \rightarrow (1) \)
\( 5x + y = 9 \quad \rightarrow (2) \)
From equation (2), we can express \( y \) in terms of \( x \):
\( y = 9 - 5x \)
Now, substitute this expression for \( y \) into equation (1):
\( 2x - 3(9 - 5x) = 7 \)
\( 2x - 27 + 15x = 7 \)
Combine the \( x \) terms:
\( 17x - 27 = 7 \)
Add 27 to both sides:
\( 17x = 7 + 27 \)
\( 17x = 34 \)
Divide by 17 to find \( x \):
\( x = \frac{34}{17} \)
\( x = 2 \)
Next, substitute the value of \( x = 2 \) back into the expression for \( y \):
\( y = 9 - 5(2) \)
\( y = 9 - 10 \)
\( y = -1 \)
So, the solution for the first pair of equations is \( x = 2 \) and \( y = -1 \). This method helps find exact values for variables in a system of equations.

(ii) The given equations are:
\( 1.5x + 0.1y = 6.2 \)
\( 3x - 0.4y = 11.2 \)
To make calculations easier, multiply both equations by 10 to remove decimals:
For the first equation:
\( 10 \times (1.5x + 0.1y) = 10 \times 6.2 \)
\( 15x + y = 62 \quad \rightarrow (1) \)
From equation (1), express \( y \) in terms of \( x \):
\( y = 62 - 15x \)
For the second equation:
\( 10 \times (3x - 0.4y) = 10 \times 11.2 \)
\( 30x - 4y = 112 \)
Divide this entire equation by 2 to simplify it further:
\( \frac{30x}{2} - \frac{4y}{2} = \frac{112}{2} \)
\( 15x - 2y = 56 \quad \rightarrow (2) \)
Now, substitute the expression for \( y \) from equation (1) into equation (2):
\( 15x - 2(62 - 15x) = 56 \)
\( 15x - 124 + 30x = 56 \)
Combine the \( x \) terms:
\( 45x - 124 = 56 \)
Add 124 to both sides:
\( 45x = 56 + 124 \)
\( 45x = 180 \)
Divide by 45 to find \( x \):
\( x = \frac{180}{45} \)
\( x = 4 \)
Finally, substitute the value of \( x = 4 \) back into the expression for \( y \):
\( y = 62 - 15(4) \)
\( y = 62 - 60 \)
\( y = 2 \)
So, the solution for the second pair of equations is \( x = 4 \) and \( y = 2 \). Multiplying by 10 helped simplify the decimal terms significantly.

(iii) The given equations are:
\( 10\% \text{ of } x + 20\% \text{ of } y = 24 \)
\( 3x - y = 20 \)
First, convert the percentages to fractions or decimals for the first equation:
\( \frac{10}{100}x + \frac{20}{100}y = 24 \)
\( \frac{x}{10} + \frac{y}{5} = 24 \)
To remove the denominators, multiply the entire equation by the least common multiple of 10 and 5, which is 10:
\( 10 \left( \frac{x}{10} + \frac{y}{5} \right) = 10 \times 24 \)
\( x + 2y = 240 \quad \rightarrow (1) \)
Now, consider the second equation:
\( 3x - y = 20 \quad \rightarrow (2) \)
From equation (2), express \( y \) in terms of \( x \):
\( -y = 20 - 3x \)
Multiply by -1 to get \( y \):
\( y = 3x - 20 \)
Substitute this expression for \( y \) into equation (1):
\( x + 2(3x - 20) = 240 \)
\( x + 6x - 40 = 240 \)
Combine the \( x \) terms:
\( 7x - 40 = 240 \)
Add 40 to both sides:
\( 7x = 240 + 40 \)
\( 7x = 280 \)
Divide by 7 to find \( x \):
\( x = \frac{280}{7} \)
\( x = 40 \)
Finally, substitute the value of \( x = 40 \) back into the expression for \( y \):
\( y = 3(40) - 20 \)
\( y = 120 - 20 \)
\( y = 100 \)
So, the solution for the third pair of equations is \( x = 40 \) and \( y = 100 \). Converting percentages to fractions is often the first step in solving such problems.

(iv) The given equations are:
\( \sqrt{2}x - \sqrt{3}y = 1 \quad \rightarrow (1) \)
\( \sqrt{3}x - \sqrt{8}y = 0 \quad \rightarrow (2) \)
From equation (1), express \( y \) in terms of \( x \):
\( -\sqrt{3}y = 1 - \sqrt{2}x \)
Multiply by -1:
\( \sqrt{3}y = \sqrt{2}x - 1 \)
Divide by \( \sqrt{3} \):
\( y = \frac{\sqrt{2}x - 1}{\sqrt{3}} \)
Now, substitute this expression for \( y \) into equation (2):
\( \sqrt{3}x - \sqrt{8}\left(\frac{\sqrt{2}x - 1}{\sqrt{3}}\right) = 0 \)
To clear the denominator \( \sqrt{3} \), multiply the entire equation by \( \sqrt{3} \):
\( \sqrt{3}(\sqrt{3}x) - \sqrt{3} \cdot \sqrt{8}\left(\frac{\sqrt{2}x - 1}{\sqrt{3}}\right) = \sqrt{3} \cdot 0 \)
\( 3x - \sqrt{8}(\sqrt{2}x - 1) = 0 \)
Expand the term \( \sqrt{8}(\sqrt{2}x - 1) \). Remember \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \).
So, \( \sqrt{8} \times \sqrt{2}x = \sqrt{16}x = 4x \). And \( \sqrt{8} \times (-1) = -\sqrt{8} \).
\( 3x - (4x - \sqrt{8}) = 0 \)
\( 3x - 4x + \sqrt{8} = 0 \)
Combine the \( x \) terms:
\( -x + \sqrt{8} = 0 \)
\( -x = -\sqrt{8} \)
\( x = \sqrt{8} \)
To simplify \( \sqrt{8} \), we can write it as \( 2\sqrt{2} \). So, \( x = 2\sqrt{2} \).
Next, substitute the value of \( x = \sqrt{8} \) back into the expression for \( y \):
\( y = \frac{\sqrt{2}(\sqrt{8}) - 1}{\sqrt{3}} \)
\( y = \frac{\sqrt{16} - 1}{\sqrt{3}} \)
\( y = \frac{4 - 1}{\sqrt{3}} \)
\( y = \frac{3}{\sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\( y = \frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( y = \frac{3\sqrt{3}}{3} \)
\( y = \sqrt{3} \)
So, the solution for the fourth pair of equations is \( x = \sqrt{8} \) (or \( 2\sqrt{2} \)) and \( y = \sqrt{3} \). Working with square roots requires careful simplification and rationalization.
In simple words: For each pair of equations, we used the substitution method. This means we solved one equation for one variable, then put that expression into the other equation to find the value of the first variable. After that, we used the found value to get the second variable. This process helps us find the exact numbers that make both equations true.

🎯 Exam Tip: Always check your final solutions by substituting the values of \( x \) and \( y \) back into both original equations. If both equations hold true, your solution is correct.

Question 2. Raman's age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Answer: Let's say Raman's current age is \( x \) years. Let the sum of his two sons' current ages be \( y \) years.
Based on the first condition, Raman's age is three times the sum of his sons' ages:
\( x = 3y \)
We can rearrange this as:
\( x - 3y = 0 \quad \rightarrow (1) \)
Now, let's look at their ages after 5 years:
Raman's age after 5 years will be \( x + 5 \) years.
Since there are two sons, each son's age will increase by 5 years. So, the sum of their ages will increase by \( 5 + 5 = 10 \) years. Therefore, the sum of his sons' ages after 5 years will be \( y + 10 \) years.
According to the second condition, after 5 years, Raman's age will be twice the sum of his sons' ages:
\( x + 5 = 2(y + 10) \)
Expand the right side:
\( x + 5 = 2y + 20 \)
Rearrange this equation:
\( x - 2y = 20 - 5 \)
\( x - 2y = 15 \quad \rightarrow (2) \)
Now we have a system of two linear equations:
1. \( x - 3y = 0 \)
2. \( x - 2y = 15 \)
From equation (1), we already know that \( x = 3y \). We can substitute this value of \( x \) into equation (2):
\( (3y) - 2y = 15 \)
\( y = 15 \)
Now that we have the value of \( y \), we can substitute it back into \( x = 3y \) to find Raman's age:
\( x = 3(15) \)
\( x = 45 \)
So, Raman's current age is 45 years. Age problems help us practice setting up and solving algebraic equations to find unknown values based on given conditions.
In simple words: Raman's age is 45 years. We found this by setting up equations based on the information given. First, Raman's age was three times his sons' combined age. Then, we looked at how their ages changed after five years and formed another equation. Solving these two equations together gave us Raman's age.

🎯 Exam Tip: When dealing with age problems involving multiple people, remember that everyone ages by the same amount. If there are two sons, their combined age increases by 5+5=10 years, not just 5.

Question 3. The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number.
Answer: Let the number be a three-digit number. We are told its middle digit is zero. So, the number can be written as \( X0Y \), where \( X \) is the hundreds digit and \( Y \) is the units digit.
The value of the original number is \( 100X + 0 \times 10 + Y = 100X + Y \).
From the first condition, the sum of the other digits (hundreds and units digits) is 13:
\( X + Y = 13 \quad \rightarrow (1) \)
Now, let's consider the number formed when the digits are reversed. If we reverse \( X0Y \), the new number becomes \( Y0X \).
The value of the reversed number is \( 100Y + 0 \times 10 + X = 100Y + X \).
From the second condition, the reversed number exceeds the original number by 495. This means:
\( 100Y + X = (100X + Y) + 495 \)
To simplify this equation, bring all terms involving \( X \) and \( Y \) to one side:
\( 100Y + X - 100X - Y = 495 \)
Combine like terms:
\( (100Y - Y) + (X - 100X) = 495 \)
\( 99Y - 99X = 495 \)
Divide the entire equation by 99 to simplify:
\( \frac{99Y}{99} - \frac{99X}{99} = \frac{495}{99} \)
\( Y - X = 5 \)
We can rewrite this as:
\( -X + Y = 5 \quad \rightarrow (2) \)
Now we have a system of two linear equations:
1. \( X + Y = 13 \)
2. \( -X + Y = 5 \)
We can solve this system by adding the two equations together. This will eliminate \( X \):
\( (X + Y) + (-X + Y) = 13 + 5 \)
\( 2Y = 18 \)
Divide by 2 to find \( Y \):
\( Y = \frac{18}{2} \)
\( Y = 9 \)
Now, substitute the value of \( Y = 9 \) back into equation (1) to find \( X \):
\( X + 9 = 13 \)
\( X = 13 - 9 \)
\( X = 4 \)
So, the hundreds digit \( X = 4 \) and the units digit \( Y = 9 \). The middle digit is 0. Therefore, the original number is \( 409 \). This problem shows how to represent numbers based on their digits and set up equations from given conditions.
In simple words: The number is 409. We figured this out by using two facts: the first and last digits add up to 13, and when the digits are swapped (making 904), the new number is 495 bigger than the original. We used these facts to make two simple math problems that helped us find the digits.

🎯 Exam Tip: When working with digit problems, remember that a number like 'abc' can be written as \( 100a + 10b + c \). This helps you form correct algebraic equations for the number's value.