Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.1

Question 1. Solve the following system of linear equations in three variables
(i) x + y + z = 5
2x - y + z = 9
x – 2y + 3z = 16
Answer:
We are given the following system of equations:
\( x + y + z = 5 \).....(1)
\( 2x - y + z = 9 \).....(2)
\( x - 2y + 3z = 16 \).....(3)

First, we add equation (1) and equation (2):
\( x+y+z=5 \)
\( 2x-y+z=9 \)
\( \implies \) \( 3x+0+2z=14 \)
So, \( 3x+2z=14 \).....(4)

Next, we multiply equation (1) by 2 and add it to equation (3):
Equation (1) multiplied by 2: \( 2(x+y+z) = 2(5) \)
\( 2x+2y+2z=10 \)
Now, we add this new equation to equation (3):
\( x-2y+3z=16 \).....(3)
\( 2x+2y+2z=10 \)
\( \implies \) \( 3x+0+5z=26 \)
So, \( 3x+5z=26 \).....(5)

Now, we have a new system with two variables, x and z. Subtract equation (4) from equation (5):
\( 3x+5z=26 \).....(5)
\( 3x+2z=14 \).....(4)
\( \implies \) \( (3x-3x) + (5z-2z) = 26-14 \)
\( \implies \) \( 0 + 3z = 12 \)
\( \implies \) \( 3z=12 \)
\( \implies \) \( z = \frac{12}{3} \)
\( \implies \) \( z = 4 \)

Next, we substitute the value of \( z=4 \) into equation (4):
\( 3x + 2z = 14 \)
\( 3x + 2(4) = 14 \)
\( 3x + 8 = 14 \)
\( 3x = 14 - 8 \)
\( 3x = 6 \)
\( x = \frac{6}{3} \)
\( x = 2 \)

Finally, we substitute the values of \( x=2 \) and \( z=4 \) into equation (1):
\( x + y + z = 5 \)
\( 2 + y + 4 = 5 \)
\( y + 6 = 5 \)
\( y = 5 - 6 \)
\( y = -1 \)

Therefore, the solution to the system is \( x=2, y=-1, \) and \( z=4 \). Finding these values means we found the single point where all three planes intersect.
In simple words: We used the given equations to find the values for x, y, and z. We combined the equations in steps to remove one variable at a time until we found all three numbers.

🎯 Exam Tip: Always verify your final answer by substituting the values of x, y, and z back into all three original equations to ensure they hold true.

Question 1.
(ii) \( \frac { 1 }{ x } – \frac { 2 }{ y } + 4 = 0, \frac { 1 }{ y } – \frac { 1 }{ z } + 1 = 0, \frac { 2 }{ z } + \frac { 3 }{ x } = 14 \)
Answer:
We introduce new variables to simplify the equations:
Let \( p = \frac { 1 }{ x }, q = \frac { 1 }{ y }, \) and \( r = \frac { 1 }{ z } \).
The given equations become:

1. \( p - 2q + 4 = 0 \)
\( \implies \) \( p - 2q = -4 \).....(1)

2. \( q - r + 1 = 0 \)
\( \implies \) \( q - r = -1 \).....(2)

3. \( \frac { 2 }{ z } + \frac { 3 }{ x } = 14 \)
\( \implies \) \( 3p + 2r = 14 \).....(3)

Now we solve this system of equations for p, q, and r. Multiply equation (2) by 2:
\( 2(q - r) = 2(-1) \)
\( 2q - 2r = -2 \)
Next, add equation (1) and this new equation:
\( p - 2q = -4 \).....(1)
\( 2q - 2r = -2 \)
\( \implies \) \( p + (2q-2q) - 2r = -4 - 2 \)
\( \implies \) \( p - 2r = -6 \).....(4)

We now have two equations with p and r: equation (3) and equation (4). Add equation (3) and equation (4):
\( 3p + 2r = 14 \).....(3)
\( p - 2r = -6 \).....(4)
\( \implies \) \( (3p+p) + (2r-2r) = 14 - 6 \)
\( \implies \) \( 4p = 8 \)
\( \implies \) \( p = \frac{8}{4} \)
\( \implies \) \( p = 2 \)

Substitute the value of \( p=2 \) into equation (1):
\( p - 2q = -4 \)
\( 2 - 2q = -4 \)
\( -2q = -4 - 2 \)
\( -2q = -6 \)
\( q = \frac{-6}{-2} \)
\( q = 3 \)

Substitute the value of \( q=3 \) into equation (2):
\( q - r = -1 \)
\( 3 - r = -1 \)
\( -r = -1 - 3 \)
\( -r = -4 \)
\( r = 4 \)

Finally, we convert back to x, y, and z using our initial substitutions:
\( p = \frac{1}{x} \implies 2 = \frac{1}{x} \implies x = \frac{1}{2} \)
\( q = \frac{1}{y} \implies 3 = \frac{1}{y} \implies y = \frac{1}{3} \)
\( r = \frac{1}{z} \implies 4 = \frac{1}{z} \implies z = \frac{1}{4} \)
The variables were introduced to simplify the initial complex fractions into a standard linear system.
In simple words: We changed the fractions into simpler letters like p, q, r. Then we solved these new equations to find p, q, and r. In the end, we converted them back to find the actual values of x, y, and z.

🎯 Exam Tip: When dealing with variables in the denominator, substitution is a powerful technique to transform non-linear equations into a solvable linear system.

Question 1.
(iii) \( x + 20 = \frac { 3y }{ 2 } + 10 = 2z + 5 = 110 – (y + z) \)
Answer:
We can break this extended equality into three separate equations:

Equation 1: \( x + 20 = \frac { 3y }{ 2 } + 10 \)
Multiply the entire equation by 2 to clear the fraction:
\( 2(x + 20) = 2(\frac { 3y }{ 2 } + 10) \)
\( 2x + 40 = 3y + 20 \)
Rearrange the terms to form a standard linear equation:
\( 2x - 3y = 20 - 40 \)
\( 2x - 3y = -20 \).....(1)

Equation 2: \( \frac { 3y }{ 2 } + 10 = 2z + 5 \)
Multiply the entire equation by 2 to clear the fraction:
\( 2(\frac { 3y }{ 2 } + 10) = 2(2z + 5) \)
\( 3y + 20 = 4z + 10 \)
Rearrange the terms:
\( 3y - 4z = 10 - 20 \)
\( 3y - 4z = -10 \).....(2)

Equation 3: \( 2z + 5 = 110 – (y + z) \)
Distribute the negative sign and simplify:
\( 2z + 5 = 110 - y - z \)
Move y and z terms to the left side and constants to the right:
\( y + 2z + z = 110 - 5 \)
\( y + 3z = 105 \).....(3)

Now we solve the system of three linear equations: (1), (2), and (3).

Multiply equation (3) by 3:
\( 3(y + 3z) = 3(105) \)
\( 3y + 9z = 315 \).....(3_new)

Subtract equation (2) from this new equation (3_new):
\( (3y + 9z) - (3y - 4z) = 315 - (-10) \)
\( 3y - 3y + 9z + 4z = 315 + 10 \)
\( 13z = 325 \)
\( z = \frac{325}{13} \)
\( z = 25 \)

Substitute the value of \( z=25 \) into equation (2):
\( 3y - 4z = -10 \)
\( 3y - 4(25) = -10 \)
\( 3y - 100 = -10 \)
\( 3y = -10 + 100 \)
\( 3y = 90 \)
\( y = \frac{90}{3} \)
\( y = 30 \)

Substitute the value of \( y=30 \) into equation (1):
\( 2x - 3y = -20 \)
\( 2x - 3(30) = -20 \)
\( 2x - 90 = -20 \)
\( 2x = -20 + 90 \)
\( 2x = 70 \)
\( x = \frac{70}{2} \)
\( x = 35 \)

So, the solution to the system is \( x=35, y=30, \) and \( z=25 \). Each step systematically eliminated a variable to find the solution.
In simple words: We took the long equation and broke it into three simpler ones. Then we solved these three equations step-by-step, finding x, y, and z one by one.

🎯 Exam Tip: When you have an equality string like A=B=C=D, remember to form a system of equations by pairing them up (e.g., A=B, B=C, C=D or A=B, A=C, A=D), choosing the simplest pairings.

Question 2. Discuss the nature of solutions of the following system of equations
(i) x + 2y − z = 6, – 3x – 2y + 5z = -12, x − 2z = 3
Answer:
Let's write down the given system of equations:
\( x + 2y - z = 6 \).....(1)
\( -3x - 2y + 5z = -12 \).....(2)
\( x - 2z = 3 \).....(3)

First, we add equation (1) and equation (2):
\( (x + 2y - z) + (-3x - 2y + 5z) = 6 + (-12) \)
\( x - 3x + 2y - 2y - z + 5z = 6 - 12 \)
\( -2x + 4z = -6 \)
Divide this new equation by -2:
\( \frac{-2x}{-2} + \frac{4z}{-2} = \frac{-6}{-2} \)
\( x - 2z = 3 \).....(4)

Now, compare equation (3) and equation (4):
Equation (3) is \( x - 2z = 3 \).
Equation (4) is \( x - 2z = 3 \).
Since equation (3) and equation (4) are identical, this means that the system has an infinite number of solutions. The existence of identical equations indicates a dependency among them.
In simple words: When we combined two of the equations, we got exactly the same equation as the third one. This means there are many answers (infinite solutions) that can make all equations true.

🎯 Exam Tip: If, during elimination, you arrive at an identity (e.g., 0=0 or one equation becomes identical to another), the system has infinitely many solutions.

Question 2.
(ii) \( 2y + z = 3(- x + 1),-x + 3y - z = -4, 3x + 2y + z = -\frac { 1 }{ 2 } \)
Answer:
Let's rewrite the given equations in standard form (Ax + By + Cz = D):

1. \( 2y + z = 3(-x + 1) \)
\( 2y + z = -3x + 3 \)
\( 3x + 2y + z = 3 \).....(1)

2. \( -x + 3y - z = -4 \)
Multiply by -1 to make the x coefficient positive:
\( x - 3y + z = 4 \).....(2)

3. \( 3x + 2y + z = -\frac { 1 }{ 2 } \)
Multiply by 2 to clear the fraction:
\( 2(3x + 2y + z) = 2(-\frac { 1 }{ 2 }) \)
\( 6x + 4y + 2z = -1 \).....(3)

Now we use elimination to find the nature of the solutions.
Subtract equation (2) from equation (1):
\( (3x + 2y + z) - (x - 3y + z) = 3 - 4 \)
\( 3x - x + 2y + 3y + z - z = -1 \)
\( 2x + 5y = -1 \).....(4)

Multiply equation (1) by 2:
\( 2(3x + 2y + z) = 2(3) \)
\( 6x + 4y + 2z = 6 \).....(1_new)

Now, subtract equation (3) from this new equation (1_new):
\( (6x + 4y + 2z) - (6x + 4y + 2z) = 6 - (-1) \)
\( 6x - 6x + 4y - 4y + 2z - 2z = 6 + 1 \)
\( 0 = 7 \)

Since 0 is not equal to 7, this is a contradiction. This means that the system of equations is inconsistent and has no solution. A system is inconsistent if no values satisfy all its equations simultaneously.
In simple words: After trying to solve the equations, we ended up with a false statement (0 = 7). This means the equations don't work together, and there is no single answer that can satisfy all of them.

🎯 Exam Tip: If, during elimination, you arrive at a false statement (e.g., 0=7), the system is inconsistent and has no solution.

Question 2.
(iii) \( \frac { y+z }{ 4 } = \frac { z+x }{ 3 } = \frac { x+y }{ 2 }, x + y + z = 27 \)
Answer:
We can write two equations from the given equalities, plus the third equation given:

1. \( \frac { y+z }{ 4 } = \frac { z+x }{ 3 } \)
Cross-multiply:
\( 3(y+z) = 4(z+x) \)
\( 3y + 3z = 4z + 4x \)
Rearrange into standard form:
\( -4x + 3y + 3z - 4z = 0 \)
\( -4x + 3y - z = 0 \)
Multiply by -1 for a positive leading coefficient:
\( 4x - 3y + z = 0 \).....(1)

2. \( \frac { z+x }{ 3 } = \frac { x+y }{ 2 } \)
Cross-multiply:
\( 2(z+x) = 3(x+y) \)
\( 2z + 2x = 3x + 3y \)
Rearrange:
\( 2x - 3x - 3y + 2z = 0 \)
\( -x - 3y + 2z = 0 \)
Multiply by -1:
\( x + 3y - 2z = 0 \).....(2)

3. The third equation is given directly:
\( x + y + z = 27 \).....(3)

Now we solve the system of equations (1), (2), and (3).

Add equation (1) and equation (2):
\( (4x - 3y + z) + (x + 3y - 2z) = 0 + 0 \)
\( 4x + x - 3y + 3y + z - 2z = 0 \)
\( 5x - z = 0 \).....(4)

Now, we use equation (3) and equation (2) to create another equation with x and z.
Multiply equation (3) by 3:
\( 3(x + y + z) = 3(27) \)
\( 3x + 3y + 3z = 81 \).....(3_new)

Subtract equation (2) from this new equation (3_new):
\( (3x + 3y + 3z) - (x + 3y - 2z) = 81 - 0 \)
\( 3x - x + 3y - 3y + 3z - (-2z) = 81 \)
\( 2x + 5z = 81 \).....(5)

Now we have a system of two equations with x and z: equation (4) and equation (5).
Equation (4): \( 5x - z = 0 \)
Equation (5): \( 2x + 5z = 81 \)

Multiply equation (4) by 5:
\( 5(5x - z) = 5(0) \)
\( 25x - 5z = 0 \).....(4_new)

Add equation (4_new) and equation (5):
\( (25x - 5z) + (2x + 5z) = 0 + 81 \)
\( 25x + 2x - 5z + 5z = 81 \)
\( 27x = 81 \)
\( x = \frac{81}{27} \)
\( x = 3 \)

Substitute the value of \( x=3 \) into equation (4):
\( 5x - z = 0 \)
\( 5(3) - z = 0 \)
\( 15 - z = 0 \)
\( z = 15 \)

Substitute the values of \( x=3 \) and \( z=15 \) into equation (3):
\( x + y + z = 27 \)
\( 3 + y + 15 = 27 \)
\( y + 18 = 27 \)
\( y = 27 - 18 \)
\( y = 9 \)

Thus, the solution is \( x=3, y=9, \) and \( z=15 \). This system has a unique solution, meaning there's only one set of values that satisfies all conditions.
In simple words: We turned the given fractions into a set of three simple equations. Then we carefully combined these equations to find unique numbers for x, y, and z that make all statements true.

🎯 Exam Tip: Always double-check your initial equation formation from complex expressions, as an error here will lead to incorrect solutions even with perfect calculations later.

Question 3. Vani, her father and her grand father have an average age of 53. One-half of her grand father's age plus one-third of her father's age plus one fourth of Vani's age is 65. If 4 years ago Vani's grandfather was four times as old as Vani then how old are they all now?
Answer:
Let Vani's current age be \( x \) years.
Let Vani's father's current age be \( y \) years.
Let Vani's grandfather's current age be \( z \) years.

From the first condition (average age is 53):
\( \frac { x+y+z }{ 3 } = 53 \)
Multiply by 3:
\( x + y + z = 159 \).....(1)

From the second condition (sum of fractional ages is 65):
\( \frac { 1 }{ 2 } z + \frac { 1 }{ 3 } y + \frac { 1 }{ 4 } x = 65 \)
To clear fractions, multiply by the least common multiple of 2, 3, and 4, which is 12:
\( 12(\frac { 1 }{ 2 } z) + 12(\frac { 1 }{ 3 } y) + 12(\frac { 1 }{ 4 } x) = 12(65) \)
\( 6z + 4y + 3x = 780 \)
Rearrange in standard form:
\( 3x + 4y + 6z = 780 \).....(2)

From the third condition (4 years ago, grandfather was four times Vani's age):
Vani's age 4 years ago: \( x-4 \)
Grandfather's age 4 years ago: \( z-4 \)
So, \( z - 4 = 4(x - 4) \)
\( z - 4 = 4x - 16 \)
Rearrange into standard form:
\( 4x - z = 16 - 4 \)
\( 4x - z = 12 \).....(3)

Now we solve the system of three linear equations (1), (2), and (3).

Multiply equation (1) by 4:
\( 4(x + y + z) = 4(159) \)
\( 4x + 4y + 4z = 636 \).....(1_new)

Subtract equation (1_new) from equation (2):
\( (3x + 4y + 6z) - (4x + 4y + 4z) = 780 - 636 \)
\( 3x - 4x + 4y - 4y + 6z - 4z = 144 \)
\( -x + 2z = 144 \).....(4)

Now we have a system with two equations involving x and z: equation (3) and equation (4).
Equation (3): \( 4x - z = 12 \)
Equation (4): \( -x + 2z = 144 \)

Multiply equation (3) by 2:
\( 2(4x - z) = 2(12) \)
\( 8x - 2z = 24 \).....(3_new)

Add equation (3_new) and equation (4):
\( (8x - 2z) + (-x + 2z) = 24 + 144 \)
\( 8x - x - 2z + 2z = 168 \)
\( 7x = 168 \)
\( x = \frac{168}{7} \)
\( x = 24 \)

Substitute the value of \( x=24 \) into equation (3):
\( 4x - z = 12 \)
\( 4(24) - z = 12 \)
\( 96 - z = 12 \)
\( -z = 12 - 96 \)
\( -z = -84 \)
\( z = 84 \)

Substitute the values of \( x=24 \) and \( z=84 \) into equation (1):
\( x + y + z = 159 \)
\( 24 + y + 84 = 159 \)
\( y + 108 = 159 \)
\( y = 159 - 108 \)
\( y = 51 \)

So, Vani's current age is 24 years, her father's age is 51 years, and her grandfather's age is 84 years. Age problems often rely on careful setting up of equations based on different time points.
In simple words: We used the information about their average age and how their ages relate at different times to set up three equations. Solving these equations helped us find each person's current age.

🎯 Exam Tip: For age-related problems, clearly define variables for current ages and carefully translate conditions for past or future ages into mathematical expressions.

Question 4. The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number?
Answer:
Let the hundreds digit be \( x \), the tens digit be \( y \), and the units digit be \( z \).
The original three-digit number can be written as \( 100x + 10y + z \).
The number with digits reversed is \( 100z + 10y + x \).

From the first condition (sum of digits is 11):
\( x + y + z = 11 \).....(1)

From the second condition (reversed number relation):
\( 100z + 10y + x = 5(100x + 10y + z) + 46 \)
\( 100z + 10y + x = 500x + 50y + 5z + 46 \)
Rearrange the terms to form an equation:
\( x - 500x + 10y - 50y + 100z - 5z = 46 \)
\( -499x - 40y + 95z = 46 \)
Multiply by -1 to get a positive leading coefficient:
\( 499x + 40y - 95z = -46 \).....(2)

From the third condition (relation between digits):
\( x + 2y = z \)
Rearrange into standard form:
\( x + 2y - z = 0 \).....(3)

Now we solve the system of equations (1), (2), and (3).

Multiply equation (1) by 95:
\( 95(x + y + z) = 95(11) \)
\( 95x + 95y + 95z = 1045 \).....(1_new)

Add equation (1_new) and equation (2):
\( (95x + 95y + 95z) + (499x + 40y - 95z) = 1045 + (-46) \)
\( (95+499)x + (95+40)y + (95-95)z = 999 \)
\( 594x + 135y = 999 \)
Divide the entire equation by 9 (since all coefficients are divisible by 9):
\( \frac{594x}{9} + \frac{135y}{9} = \frac{999}{9} \)
\( 66x + 15y = 111 \).....(4)

Next, add equation (1) and equation (3):
\( (x + y + z) + (x + 2y - z) = 11 + 0 \)
\( x + x + y + 2y + z - z = 11 \)
\( 2x + 3y = 11 \).....(5)

Now we have a system of two equations with x and y: equation (4) and equation (5).
Equation (4): \( 66x + 15y = 111 \)
Equation (5): \( 2x + 3y = 11 \)

To simplify equation (4), divide by 3:
\( \frac{66x}{3} + \frac{15y}{3} = \frac{111}{3} \)
\( 22x + 5y = 37 \).....(4_new)

Multiply equation (5) by 11:
\( 11(2x + 3y) = 11(11) \)
\( 22x + 33y = 121 \).....(5_new)

Subtract equation (4_new) from equation (5_new):
\( (22x + 33y) - (22x + 5y) = 121 - 37 \)
\( 22x - 22x + 33y - 5y = 84 \)
\( 28y = 84 \)
\( y = \frac{84}{28} \)
\( y = 3 \)

Substitute the value of \( y=3 \) into equation (5):
\( 2x + 3y = 11 \)
\( 2x + 3(3) = 11 \)
\( 2x + 9 = 11 \)
\( 2x = 11 - 9 \)
\( 2x = 2 \)
\( x = \frac{2}{2} \)
\( x = 1 \)

Substitute the values of \( x=1 \) and \( y=3 \) into equation (1):
\( x + y + z = 11 \)
\( 1 + 3 + z = 11 \)
\( 4 + z = 11 \)
\( z = 11 - 4 \)
\( z = 7 \)

The digits are \( x=1, y=3, \) and \( z=7 \).
Therefore, the original three-digit number is \( 100x + 10y + z = 100(1) + 10(3) + 7 = 100 + 30 + 7 = 137 \). Problems involving digits often require careful representation of the number itself.
In simple words: We wrote the three conditions given about the number's digits as mathematical equations. Then we solved these equations to find each digit, and put them together to get the original number.

🎯 Exam Tip: Remember that a three-digit number with digits x, y, z is represented as 100x + 10y + z, not just xyz. This is crucial for setting up equations correctly.

Question 5. There are 12 pieces of five, ten and twenty rupee currencies whose total value is. 2 sorts are interchanged in their numbers its value will be increased by Rs 20. Find the number of currencies in each sort.
Answer:
Let the number of Rs 5 currencies be \( x \).
Let the number of Rs 10 currencies be \( y \).
Let the number of Rs 20 currencies be \( z \).

From the first condition (total 12 pieces):
\( x + y + z = 12 \).....(1)

From the initial value (implied by solution's use of 105):
The total value of the currencies is \( 5x + 10y + 20z \). The solution implies this value is Rs 105.
So, \( 5x + 10y + 20z = 105 \)
Divide the entire equation by 5:
\( \frac{5x}{5} + \frac{10y}{5} + \frac{20z}{5} = \frac{105}{5} \)
\( x + 2y + 4z = 21 \).....(2)

From the third condition (interchanging Rs 5 and Rs 10 notes increases value by Rs 20):
If the number of Rs 5 notes (x) and Rs 10 notes (y) are interchanged, the new value becomes \( 10x + 5y + 20z \).
The original value was Rs 105, and it increases by Rs 20, so the new total value is \( 105 + 20 = 125 \).
Therefore, \( 10x + 5y + 20z = 125 \)
Divide the entire equation by 5:
\( \frac{10x}{5} + \frac{5y}{5} + \frac{20z}{5} = \frac{125}{5} \)
\( 2x + y + 4z = 25 \).....(3)

Now we solve the system of equations (1), (2), and (3).

Subtract equation (1) from equation (2):
\( (x + 2y + 4z) - (x + y + z) = 21 - 12 \)
\( x - x + 2y - y + 4z - z = 9 \)
\( y + 3z = 9 \).....(4)

Subtract equation (2) from equation (3):
\( (2x + y + 4z) - (x + 2y + 4z) = 25 - 21 \)
\( 2x - x + y - 2y + 4z - 4z = 4 \)
\( x - y = 4 \).....(5)

Now we have a system of two equations with x and y: equation (4) and equation (5).
Equation (4): \( y + 3z = 9 \)
Equation (5): \( x - y = 4 \)
The source calculation used equations (4) and (5) in a slightly different way (from an earlier step not shown in this specific OCR for Q5 but implicitly from Q1(i) similar steps), which is why some steps might look different from a direct sequence here. Let's follow the source's approach of generating new combined equations.

Let's follow the source's method on page 12 by using (1_new), (2), (3) differently:
From equation (1) and equation (2):
Multiply (1) by 4: \( 4x + 4y + 4z = 48 \).....(1_new)
Subtract (2) from (1_new):
\( (4x + 4y + 4z) - (x + 2y + 4z) = 48 - 21 \)
\( 3x + 2y = 27 \).....(4_from_source)

From equation (2) and equation (3):
Subtract (2) from (3):
\( (2x + y + 4z) - (x + 2y + 4z) = 25 - 21 \)
\( x - y = 4 \).....(5_from_source)

Now we solve the system with (4_from_source) and (5_from_source):
\( 3x + 2y = 27 \)
\( x - y = 4 \)

Multiply the second equation (\( x - y = 4 \)) by 2:
\( 2(x - y) = 2(4) \)
\( 2x - 2y = 8 \).....(5_new)

Add \( (4\_from\_source) \) and \( (5\_new) \):
\( (3x + 2y) + (2x - 2y) = 27 + 8 \)
\( 5x = 35 \)
\( x = \frac{35}{5} \)
\( x = 7 \)

Substitute the value of \( x=7 \) into equation (5_from_source):
\( x - y = 4 \)
\( 7 - y = 4 \)
\( -y = 4 - 7 \)
\( -y = -3 \)
\( y = 3 \)

Substitute the values of \( x=7 \) and \( y=3 \) into equation (1):
\( x + y + z = 12 \)
\( 7 + 3 + z = 12 \)
\( 10 + z = 12 \)
\( z = 12 - 10 \)
\( z = 2 \)

The number of Rs 5 currencies is 7.
The number of Rs 10 currencies is 3.
The number of Rs 20 currencies is 2.
This problem shows how systems of equations can model real-world scenarios involving currency and value.
In simple words: We made three equations based on the total number of notes, their total value, and what happens when we swap two types of notes. Solving these equations helped us find how many of each currency note there were.

🎯 Exam Tip: For currency-related problems, clearly define variables for the number of each denomination. Remember to form equations for both the total count of items and their total monetary value.