Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.12

Question 1. Solve by the method of elimination
(i) \( 2x - y = 3; 3x + y = 7 \)
(ii) \( x - y = 5; 3x + 2y = 25 \)
(iii) \( \frac{x}{10} + \frac{y}{5} = 14; \frac{x}{8} + \frac{y}{6} = 15 \)
(iv) \( 3(2x + y) = 7xy; 3(x + 3y) = 11xy \)
(v) \( \frac{4}{x} + 5y = 7; \frac{3}{x} + 4y = 5 \)
(vi) \( 13x + 11y = 70; 11x + 13y = 74 \)
Answer:
(i) Given the equations:
\( 2x - y = 3 \rightarrow (1) \)
\( 3x + y = 7 \rightarrow (2) \)
Now, we add equation (1) and equation (2):
\( (2x - y) + (3x + y) = 3 + 7 \)
\( 5x + 0 = 10 \)
\( 5x = 10 \)
\( x = \frac{10}{5} \)
\( x = 2 \)
Next, substitute the value of \( x = 2 \) into equation (1):
\( 2(2) - y = 3 \)
\( 4 - y = 3 \)
\( -y = 3 - 4 \)
\( -y = -1 \)
\( y = 1 \)
So, the values are \( x = 2 \) and \( y = 1 \). This method quickly solves systems where one variable's coefficients are opposites.
In simple words: We have two equations. We add them together to remove \( y \), which helps us find \( x \). Then, we put the value of \( x \) back into one equation to find \( y \).

🎯 Exam Tip: Always verify your solution by substituting the calculated \( x \) and \( y \) values into both original equations to ensure they are satisfied.

Answer:
(ii) Given the equations:
\( x - y = 5 \rightarrow (1) \)
\( 3x + 2y = 25 \rightarrow (2) \)
To eliminate \( y \), we multiply equation (1) by 2:
\( (1) \times 2 \implies 2(x - y) = 2(5) \)
\( 2x - 2y = 10 \rightarrow (3) \)
Now, we multiply equation (2) by 1 (to keep it as is):
\( (2) \times 1 \implies 3x + 2y = 25 \rightarrow (2) \)
Next, we add equation (3) and equation (2):
\( (2x - 2y) + (3x + 2y) = 10 + 25 \)
\( 5x + 0 = 35 \)
\( 5x = 35 \)
\( x = \frac{35}{5} \)
\( x = 7 \)
Substitute the value of \( x = 7 \) into equation (1):
\( 7 - y = 5 \)
\( -y = 5 - 7 \)
\( -y = -2 \)
\( y = 2 \)
Therefore, the values are \( x = 7 \) and \( y = 2 \). The goal of elimination is to make the coefficients of one variable the same or opposite so they cancel out.
In simple words: We changed the first equation so that the \( y \) parts would cancel out when added to the second equation. This helped us find \( x \), and then we found \( y \).

🎯 Exam Tip: Remember to multiply *all* terms in an equation by the chosen constant, including the term on the right side of the equality.

Answer:
(iii) Given the equations:
\( \frac{x}{10} + \frac{y}{5} = 14 \)
First, find the Least Common Multiple (LCM) of the denominators 10 and 5, which is 10. Multiply the entire equation by 10 to remove fractions:
\( 10(\frac{x}{10}) + 10(\frac{y}{5}) = 10(14) \)
\( x + 2y = 140 \rightarrow (1) \)
Next, consider the second equation:
\( \frac{x}{8} + \frac{y}{6} = 15 \)
Find the LCM of the denominators 8 and 6, which is 24. Multiply the entire equation by 24 to remove fractions:
\( 24(\frac{x}{8}) + 24(\frac{y}{6}) = 24(15) \)
\( 3x + 4y = 360 \rightarrow (2) \)
Now we have a system of simpler equations:
\( x + 2y = 140 \rightarrow (1) \)
\( 3x + 4y = 360 \rightarrow (2) \)
To eliminate \( y \), multiply equation (1) by 2:
\( (1) \times 2 \implies 2(x + 2y) = 2(140) \)
\( 2x + 4y = 280 \rightarrow (3) \)
Multiply equation (2) by 1 (keep it as is):
\( (2) \times 1 \implies 3x + 4y = 360 \rightarrow (2) \)
Subtract equation (2) from equation (3):
\( (2x + 4y) - (3x + 4y) = 280 - 360 \)
\( -x + 0 = -80 \)
\( -x = -80 \)
\( x = 80 \)
Substitute the value of \( x = 80 \) into equation (1):
\( 80 + 2y = 140 \)
\( 2y = 140 - 80 \)
\( 2y = 60 \)
\( y = \frac{60}{2} \)
\( y = 30 \)
Thus, the values are \( x = 80 \) and \( y = 30 \). Clearing the fractions first makes solving the system much easier.
In simple words: We first got rid of the fractions in both equations by multiplying by the smallest common number. Then we used the elimination method to find \( x \) and \( y \), just like before.

🎯 Exam Tip: Always remove fractions by multiplying by the LCM of denominators at the beginning to simplify the equations into standard linear forms.

Answer:
(iv) Given the equations:
\( 3(2x + y) = 7xy \)
\( 6x + 3y = 7xy \)
To solve this, we divide the entire equation by \( xy \). This is allowed if \( x, y \ne 0 \).
\( \frac{6x}{xy} + \frac{3y}{xy} = \frac{7xy}{xy} \)
\( \frac{6}{y} + \frac{3}{x} = 7 \)
Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \). Substituting these into the equation:
\( 6b + 3a = 7 \rightarrow (1) \)
Now, for the second equation:
\( 3(x + 3y) = 11xy \)
\( 3x + 9y = 11xy \)
Divide the entire equation by \( xy \):
\( \frac{3x}{xy} + \frac{9y}{xy} = \frac{11xy}{xy} \)
\( \frac{3}{y} + \frac{9}{x} = 11 \)
Substitute \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \):
\( 3b + 9a = 11 \rightarrow (2) \)
Rearranging equations (1) and (2) in terms of \( a \) and \( b \):
\( 3a + 6b = 7 \rightarrow (1) \)
\( 9a + 3b = 11 \rightarrow (2) \)
To eliminate \( a \), multiply equation (1) by 3:
\( (1) \times 3 \implies 3(3a + 6b) = 3(7) \)
\( 9a + 18b = 21 \rightarrow (3) \)
Multiply equation (2) by 1 (keep as is):
\( (2) \times 1 \implies 9a + 3b = 11 \rightarrow (2) \)
Subtract equation (2) from equation (3):
\( (9a + 18b) - (9a + 3b) = 21 - 11 \)
\( 0 + 15b = 10 \)
\( 15b = 10 \)
\( b = \frac{10}{15} \)
\( b = \frac{2}{3} \)
Substitute the value of \( b = \frac{2}{3} \) into equation (1):
\( 3a + 6(\frac{2}{3}) = 7 \)
\( 3a + 4 = 7 \)
\( 3a = 7 - 4 \)
\( 3a = 3 \)
\( a = \frac{3}{3} \)
\( a = 1 \)
Now, we convert back from \( a \) and \( b \) to \( x \) and \( y \):
Since \( a = \frac{1}{x} \), then \( 1 = \frac{1}{x} \implies x = 1 \)
Since \( b = \frac{1}{y} \), then \( \frac{2}{3} = \frac{1}{y} \implies y = \frac{3}{2} \)
Therefore, the values are \( x = 1 \) and \( y = \frac{3}{2} \). Using substitution for reciprocal terms is a common technique to linearize such equations.
In simple words: These equations had \( xy \) terms, so we divided everything by \( xy \) and then used new letters (\( a \) and \( b \)) for \( \frac{1}{x} \) and \( \frac{1}{y} \). We solved for \( a \) and \( b \), and then changed them back to find \( x \) and \( y \).

🎯 Exam Tip: When equations involve product terms like \( xy \), try dividing by \( xy \) and then substituting \( \frac{1}{x} \) and \( \frac{1}{y} \) to convert them into linear equations.

Answer:
(v) Given the equations:
\( \frac{4}{x} + 5y = 7 \)
\( \frac{3}{x} + 4y = 5 \)
Let \( \frac{1}{x} = a \). Substituting this into the equations:
\( 4a + 5y = 7 \rightarrow (1) \)
\( 3a + 4y = 5 \rightarrow (2) \)
To eliminate \( y \), multiply equation (1) by 4 and equation (2) by 5:
\( (1) \times 4 \implies 4(4a + 5y) = 4(7) \)
\( 16a + 20y = 28 \rightarrow (3) \)
\( (2) \times 5 \implies 5(3a + 4y) = 5(5) \)
\( 15a + 20y = 25 \rightarrow (4) \)
Subtract equation (4) from equation (3):
\( (16a + 20y) - (15a + 20y) = 28 - 25 \)
\( a + 0 = 3 \)
\( a = 3 \)
Substitute the value of \( a = 3 \) into equation (1):
\( 4(3) + 5y = 7 \)
\( 12 + 5y = 7 \)
\( 5y = 7 - 12 \)
\( 5y = -5 \)
\( y = \frac{-5}{5} \)
\( y = -1 \)
Now, convert back from \( a \) to \( x \):
Since \( a = \frac{1}{x} \), then \( 3 = \frac{1}{x} \implies 3x = 1 \implies x = \frac{1}{3} \)
Therefore, the values are \( x = \frac{1}{3} \) and \( y = -1 \). This method simplifies the problem into a standard linear system.
In simple words: We replaced \( \frac{1}{x} \) with a new letter, \( a \), to make the equations simpler. Then we solved for \( a \) and \( y \) using elimination. Finally, we used the value of \( a \) to find the real value of \( x \).

🎯 Exam Tip: When a variable appears in the denominator, substitute a new variable (e.g., \( a = \frac{1}{x} \)) to transform the equations into linear form, making them easier to solve.

Answer:
(vi) Given the equations:
\( 13x + 11y = 70 \rightarrow (1) \)
\( 11x + 13y = 74 \rightarrow (2) \)
When coefficients of \( x \) and \( y \) are swapped between two equations, we can add and subtract them. First, add equation (1) and equation (2):
\( (13x + 11y) + (11x + 13y) = 70 + 74 \)
\( 24x + 24y = 144 \)
Divide the entire equation by 24:
\( \frac{24x}{24} + \frac{24y}{24} = \frac{144}{24} \)
\( x + y = 6 \rightarrow (3) \)
Next, subtract equation (2) from equation (1):
\( (13x + 11y) - (11x + 13y) = 70 - 74 \)
\( 13x + 11y - 11x - 13y = -4 \)
\( 2x - 2y = -4 \)
Divide the entire equation by 2:
\( \frac{2x}{2} - \frac{2y}{2} = \frac{-4}{2} \)
\( x - y = -2 \rightarrow (4) \)
Now we have a simpler system of equations:
\( x + y = 6 \rightarrow (3) \)
\( x - y = -2 \rightarrow (4) \)
Add equation (3) and equation (4):
\( (x + y) + (x - y) = 6 + (-2) \)
\( 2x + 0 = 4 \)
\( 2x = 4 \)
\( x = \frac{4}{2} \)
\( x = 2 \)
Substitute the value of \( x = 2 \) into equation (3):
\( 2 + y = 6 \)
\( y = 6 - 2 \)
\( y = 4 \)
Therefore, the values are \( x = 2 \) and \( y = 4 \). This method is especially efficient for these types of symmetric coefficient problems.
In simple words: We first added the two equations, then subtracted them to get two new, simpler equations. Then, we solved these new equations to find \( x \) and \( y \).

🎯 Exam Tip: For equations where coefficients of \( x \) and \( y \) are swapped (e.g., \( ax + by \) and \( bx + ay \)), a quick approach is to add them once and subtract them once to get two simpler linear equations.

Question 2. The monthly income of A and B are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 5,000 per month, find the monthly income of each.
Answer: Let the monthly income of person A be \( x \) and the monthly income of person B be \( y \).
According to the first condition, the ratio of their incomes is 3:4:
\( \frac{x}{y} = \frac{3}{4} \)
\( 4x = 3y \)
\( 4x - 3y = 0 \rightarrow (1) \)
Now, let's consider their expenditures. Each person saves Rs 5,000 per month. Savings are calculated as income minus expenditure. So, expenditure is income minus savings.
Expenditure of A \( = x - \text{Rs } 5,000 \)
Expenditure of B \( = y - \text{Rs } 5,000 \)
According to the second condition, the ratio of their expenditures is 5:7:
\( \frac{x - 5000}{y - 5000} = \frac{5}{7} \)
\( 7(x - 5000) = 5(y - 5000) \)
\( 7x - 35000 = 5y - 25000 \)
\( 7x - 5y = -25000 + 35000 \)
\( 7x - 5y = 10000 \rightarrow (2) \)
Now we have a system of two linear equations:
\( 4x - 3y = 0 \rightarrow (1) \)
\( 7x - 5y = 10000 \rightarrow (2) \)
To eliminate \( y \), multiply equation (1) by 5 and equation (2) by 3:
\( (1) \times 5 \implies 5(4x - 3y) = 5(0) \)
\( 20x - 15y = 0 \rightarrow (3) \)
\( (2) \times 3 \implies 3(7x - 5y) = 3(10000) \)
\( 21x - 15y = 30000 \rightarrow (4) \)
Subtract equation (3) from equation (4):
\( (21x - 15y) - (20x - 15y) = 30000 - 0 \)
\( x + 0 = 30000 \)
\( x = 30000 \)
Substitute the value of \( x = 30000 \) into equation (1):
\( 4(30000) - 3y = 0 \)
\( 120000 = 3y \)
\( y = \frac{120000}{3} \)
\( y = 40000 \)
Therefore, the monthly income of A is Rs 30,000 and the monthly income of B is Rs 40,000. Savings are simply the difference between what one earns and what one spends.
In simple words: We used the income and expenditure ratios to make two equations. Since both people saved the same amount, we could write their expenses based on their income. Solving these equations told us how much money each person earns every month.

🎯 Exam Tip: When dealing with ratio problems, always represent the quantities as \( kx \) and \( ky \) (or similar) to form the initial equations correctly.

Question 3. Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Answer: Let the present age of the man be \( x \) years and the present age of his son be \( y \) years.
First, consider the situation five years ago:
Age of the man five years ago \( = x - 5 \) years
Age of his son five years ago \( = y - 5 \) years
According to the first condition, five years ago, the man was seven times as old as his son:
\( x - 5 = 7(y - 5) \)
\( x - 5 = 7y - 35 \)
\( x - 7y = -35 + 5 \)
\( x - 7y = -30 \rightarrow (1) \)
Next, consider the situation five years from now (five years hence):
Age of the man five years from now \( = x + 5 \) years
Age of his son five years from now \( = y + 5 \) years
According to the second condition, five years hence, the man will be four times as old as his son:
\( x + 5 = 4(y + 5) \)
\( x + 5 = 4y + 20 \)
\( x - 4y = 20 - 5 \)
\( x - 4y = 15 \rightarrow (2) \)
Now we have a system of two linear equations:
\( x - 7y = -30 \rightarrow (1) \)
\( x - 4y = 15 \rightarrow (2) \)
To eliminate \( x \), subtract equation (2) from equation (1):
\( (x - 7y) - (x - 4y) = -30 - 15 \)
\( x - 7y - x + 4y = -45 \)
\( -3y = -45 \)
\( y = \frac{-45}{-3} \)
\( y = 15 \)
Substitute the value of \( y = 15 \) into equation (1):
\( x - 7(15) = -30 \)
\( x - 105 = -30 \)
\( x = -30 + 105 \)
\( x = 75 \)
Therefore, the present age of the man is 75 years and the present age of his son is 15 years. Age problems require careful setup of variables for past, present, and future scenarios.
In simple words: We set up equations for the man and his son's ages based on what happened five years ago and what will happen five years from now. Then we solved these equations to find their current ages.

🎯 Exam Tip: For age problems, always define variables for *present* ages first, and then express past or future ages relative to these present ages.