Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.13

Question 1. Solve by cross-multiplication method.
(i) \( 8x - 3y = 12; 5x = 2y + 7 \)
Answer:
First, we need to write the given equations in the standard form \( ax + by + c = 0 \).
The equations are:
\( 8x - 3y - 12 = 0 \) → (1)
\( 5x - 2y - 7 = 0 \) → (2)
Now, we use the coefficients for cross-multiplication. This method helps solve systems of linear equations efficiently.

\( \frac{x}{(-3)(-7) - (-12)(-2)} = \frac{y}{(-12)(5) - (8)(-7)} = \frac{1}{(8)(-2) - (-3)(5)} \)
\( \frac{x}{21 - 24} = \frac{y}{-60 + 56} = \frac{1}{-16 + 15} \)
\( \frac{x}{-3} = \frac{y}{-4} = \frac{1}{-1} \)
To find \( x \), we set the first and third parts equal:
\( \frac{x}{-3} = -1 \)
\( \implies x = (-1) \times (-3) \)
\( \implies x = 3 \)
To find \( y \), we set the second and third parts equal:
\( \frac{y}{-4} = -1 \)
\( \implies y = (-1) \times (-4) \)
\( \implies y = 4 \)
So, the value of \( x = 3 \) and \( y = 4 \). This method provides a systematic way to solve for variables in linear equations.
In simple words: First, arrange the equations. Then, use a special multiplication rule with the numbers in front of \( x, y \) and the constants to find \( x \) and \( y \) quickly.

🎯 Exam Tip: Always convert equations to the standard form \( ax + by + c = 0 \) before applying the cross-multiplication method to avoid sign errors.

Question 1. Solve by cross-multiplication method.
(ii) \( 6x + 7y - 11 = 0; 5x + 2y = 13 \)
Answer:
First, we need to write the given equations in the standard form \( ax + by + c = 0 \).
The equations are:
\( 6x + 7y - 11 = 0 \) → (1)
\( 5x + 2y - 13 = 0 \) → (2)
Now, we use the coefficients for cross-multiplication.

\( \frac{x}{(7)(-13) - (-11)(2)} = \frac{y}{(-11)(5) - (6)(-13)} = \frac{1}{(6)(2) - (7)(5)} \)
\( \frac{x}{-91 - (-22)} = \frac{y}{-55 - (-78)} = \frac{1}{12 - 35} \)
\( \frac{x}{-91 + 22} = \frac{y}{-55 + 78} = \frac{1}{-23} \)
\( \frac{x}{-69} = \frac{y}{23} = \frac{1}{-23} \)
To find \( x \), we set the first and third parts equal:
\( \frac{x}{-69} = \frac{1}{-23} \)
\( \implies -23x = -69 \)
\( \implies x = \frac{-69}{-23} \)
\( \implies x = 3 \)
To find \( y \), we set the second and third parts equal:
\( \frac{y}{23} = \frac{1}{-23} \)
\( \implies -23y = 23 \)
\( \implies y = \frac{23}{-23} \)
\( \implies y = -1 \)
Therefore, the value of \( x = 3 \) and \( y = -1 \). This consistent approach helps in avoiding errors in complex calculations.
In simple words: Arrange equations, then use the cross-multiplication pattern with the numbers. This helps you find the values for \( x \) and \( y \) step-by-step.

🎯 Exam Tip: Pay close attention to negative signs during multiplication and subtraction steps, as a single sign error can lead to incorrect results.

Question 1. Solve by cross-multiplication method.
(iii) \( \frac{2}{x} + \frac{3}{y} = 5; \frac{3}{x} - \frac{1}{y} + 9 = 0 \)
Answer:
First, we need to introduce a substitution to convert these equations into a linear form. Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \).
Now, the given equations become:
\( 2a + 3b - 5 = 0 \) → (1)
\( 3a - b + 9 = 0 \) → (2)
We use the coefficients for cross-multiplication for these new linear equations.

\( \frac{a}{(3)(9) - (-5)(-1)} = \frac{b}{(-5)(3) - (2)(9)} = \frac{1}{(2)(-1) - (3)(3)} \)
\( \frac{a}{27 - 5} = \frac{b}{-15 - 18} = \frac{1}{-2 - 9} \)
\( \frac{a}{22} = \frac{b}{-33} = \frac{1}{-11} \)
To find \( a \), we set the first and third parts equal:
\( \frac{a}{22} = \frac{1}{-11} \)
\( \implies -11a = 22 \)
\( \implies a = \frac{22}{-11} \)
\( \implies a = -2 \)
To find \( b \), we set the second and third parts equal:
\( \frac{b}{-33} = \frac{1}{-11} \)
\( \implies -11b = -33 \)
\( \implies b = \frac{-33}{-11} \)
\( \implies b = 3 \)
Now, we substitute back the original variables:
Since \( \frac{1}{x} = a \), we have \( \frac{1}{x} = -2 \).
\( \implies -2x = 1 \)
\( \implies x = -\frac{1}{2} \)
Since \( \frac{1}{y} = b \), we have \( \frac{1}{y} = 3 \).
\( \implies 3y = 1 \)
\( \implies y = \frac{1}{3} \)
So, the value of \( x = -\frac{1}{2} \) and \( y = \frac{1}{3} \). Using substitution simplifies the initial non-linear system into a solvable linear one.
In simple words: When \( x \) and \( y \) are at the bottom of a fraction, use a trick: replace \( 1/x \) with 'a' and \( 1/y \) with 'b'. Solve for 'a' and 'b' first, then put the original \( x \) and \( y \) back to find their final values.

🎯 Exam Tip: When dealing with fractional variables, always use substitution to simplify the system into a linear form before applying methods like cross-multiplication.

Question 2. Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins totalling Rs 220, how many coins of each kind does she have.
Answer:
Let's define the variables:
Let \( x \) be the number of 2 rupee coins.
Let \( y \) be the number of 5 rupee coins.
Based on the first condition, Akshaya has 80 coins in total:
\( x + y = 80 \) → (1)
Based on the second condition, the total value of the coins is Rs 220:
\( 2x + 5y = 220 \) → (2)
Now, we convert these equations into the standard form \( ax + by + c = 0 \):
\( x + y - 80 = 0 \) → (3)
\( 2x + 5y - 220 = 0 \) → (4)
We use the coefficients for cross-multiplication to solve for \( x \) and \( y \). This method is useful for systems of linear equations involving quantities and values.

\( \frac{x}{(1)(-220) - (-80)(5)} = \frac{y}{(-80)(2) - (1)(-220)} = \frac{1}{(1)(5) - (1)(2)} \)
\( \frac{x}{-220 - (-400)} = \frac{y}{-160 - (-220)} = \frac{1}{5 - 2} \)
\( \frac{x}{-220 + 400} = \frac{y}{-160 + 220} = \frac{1}{3} \)
\( \frac{x}{180} = \frac{y}{60} = \frac{1}{3} \)
To find \( x \), we set the first and third parts equal:
\( \frac{x}{180} = \frac{1}{3} \)
\( \implies 3x = 180 \)
\( \implies x = \frac{180}{3} \)
\( \implies x = 60 \)
To find \( y \), we set the second and third parts equal:
\( \frac{y}{60} = \frac{1}{3} \)
\( \implies 3y = 60 \)
\( \implies y = \frac{60}{3} \)
\( \implies y = 20 \)
Therefore, Akshaya has 60 coins of 2 rupees and 20 coins of 5 rupees. These word problems require careful translation into algebraic equations before solving.
In simple words: We made two equations: one for the total number of coins and one for their total money value. By solving these, we found out how many 2-rupee coins and how many 5-rupee coins Akshaya has.

🎯 Exam Tip: Clearly define your variables for each unknown quantity. This makes it easier to set up the correct equations from the problem's conditions.

Question 3. It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half of the pool is filled. How long would each pipe take to fill the swimming pool.
Answer:
Let's define the variables:
Let \( x \) be the time (in hours) the larger diameter pipe takes to fill the entire pool alone.
Let \( y \) be the time (in hours) the smaller diameter pipe takes to fill the entire pool alone.
The rate of work for the larger pipe is \( \frac{1}{x} \) (fraction of pool filled per hour).
The rate of work for the smaller pipe is \( \frac{1}{y} \) (fraction of pool filled per hour).
Based on the first condition, both pipes working together fill the pool in 24 hours:
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{24} \) → (1)
Based on the second condition, if the larger pipe works for 8 hours and the smaller pipe for 18 hours, half the pool is filled:
\( \frac{8}{x} + \frac{18}{y} = \frac{1}{2} \) → (2)
To solve these equations, we introduce substitution: Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \).
The equations become:
\( a + b = \frac{1}{24} \)
\( 8a + 18b = \frac{1}{2} \)
Convert these to standard form \( ax + by + c = 0 \) to prepare for cross-multiplication:
From the first equation, multiply by 24:
\( 24a + 24b = 1 \)
\( 24a + 24b - 1 = 0 \) → (3)
From the second equation, multiply by 2:
\( 16a + 36b = 1 \)
\( 16a + 36b - 1 = 0 \) → (4)
We now use the coefficients for cross-multiplication to solve for \( a \) and \( b \). This is a common strategy for solving work-rate problems.

\( \frac{a}{(24)(-1) - (-1)(36)} = \frac{b}{(-1)(16) - (24)(-1)} = \frac{1}{(24)(36) - (24)(16)} \)
\( \frac{a}{-24 - (-36)} = \frac{b}{-16 - (-24)} = \frac{1}{864 - 384} \)
\( \frac{a}{-24 + 36} = \frac{b}{-16 + 24} = \frac{1}{480} \)
\( \frac{a}{12} = \frac{b}{8} = \frac{1}{480} \)
To find \( a \), we set the first and third parts equal:
\( \frac{a}{12} = \frac{1}{480} \)
\( \implies 480a = 12 \)
\( \implies a = \frac{12}{480} \)
\( \implies a = \frac{1}{40} \)
To find \( b \), we set the second and third parts equal:
\( \frac{b}{8} = \frac{1}{480} \)
\( \implies 480b = 8 \)
\( \implies b = \frac{8}{480} \)
\( \implies b = \frac{1}{60} \)
Now, we substitute back the original variables:
Since \( \frac{1}{x} = a \), we have \( \frac{1}{x} = \frac{1}{40} \).
\( \implies x = 40 \) hours
Since \( \frac{1}{y} = b \), we have \( \frac{1}{y} = \frac{1}{60} \).
\( \implies y = 60 \) hours
The larger pipe takes 40 hours to fill the pool alone, and the smaller pipe takes 60 hours to fill the pool alone. Understanding the concept of work rate as the inverse of time is crucial for these problems.
In simple words: We found out how fast each pipe fills the pool by itself. The larger pipe fills it in 40 hours, and the smaller pipe takes 60 hours. This was done by making equations about their combined work and solving them.

🎯 Exam Tip: For work-rate problems, remember that "rate" is usually \( \frac{1}{\text{time}} \). Convert fractions involving variables into linear equations using substitution to simplify the solving process.