Maharashtra Board Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.1 Solutions

Question 1. Complete the following activity to solve the simultaneous equations.
\( 5x + 3y = 9 \) ...(i)
\( 2x - 3y = 12 \) ...(ii)
Answer:
Let's add equations (i) and (ii).
\( 5x + 3y = 9 \)
\( + \underline{2x - 3y = 12} \)
\( 7x = 21 \)

\( \implies x = \frac{21}{7} \)

\( \implies x = 3 \)

Place \( x = 3 \) in equation (i).
\( 5(3) + 3y = 9 \)

\( \implies 15 + 3y = 9 \)

\( \implies 3y = 9 - 15 \)

\( \implies 3y = -6 \)

\( \implies y = \frac{-6}{3} \)

\( \implies y = -2 \)

Therefore, the solution is \( (x, y) = (3, -2) \).
In simple words: We add the two equations together because the \( +3y \) and \( -3y \) cancel each other out. This helps us find that \( x \) is 3, which we then use to find that \( y \) is -2.

🎯 Exam Tip: Always write down the final solution in the coordinate format \( (x, y) \) at the end of the answer to secure full marks.

Question 1. Complete the following activity to solve the simultaneous equations.
\( 5x + 3y = 9 \) ... (1)
\( 2x - 3y = 12 \) ... (2)
Answer: Let us add equations (1) and (2):
\( (5x + 3y) + (2x - 3y) = 9 + 12 \)
\( \implies 7x = 21 \)
\( \implies x = \frac{21}{7} \)
\( \implies x = 3 \)
Place \( x = 3 \) in equation (1):
\( 5 \times 3 + 3y = 9 \)
\( \implies 15 + 3y = 9 \)
\( \implies 3y = 9 - 15 \)
\( \implies 3y = -6 \)
\( \implies y = \frac{-6}{3} \)
\( \implies y = -2 \)
Therefore, the solution is \( (x, y) = (3, -2) \). This systematic method of elimination helps us find the unique point where both lines intersect.
In simple words: To solve these equations, we add them together to cancel out the y terms and find x. Once we know x, we plug it back in to find y.

🎯 Exam Tip: When adding equations to eliminate a variable, make sure the coefficients of that variable are equal in magnitude but opposite in sign.

Question 2. Solve the following simultaneous equations.
(i) \( 3a + 5b = 26 \); \( a + 5b = 22 \)
(ii) \( x + 7y = 10 \); \( 3x - 2y = 7 \)
(iii) \( 2x - 3y = 9 \); \( 2x + y = 13 \)
(iv) \( 5m - 3n = 19 \); \( m - 6n = -7 \)
(v) \( 5x + 2y = -3 \); \( x + 5y = 4 \)
(vi) \( \frac{1}{3}x + y = \frac{10}{3} \); \( 2x + \frac{1}{4}y = \frac{11}{4} \)
(vii) \( 99x + 101y = 499 \); \( 101x + 99y = 501 \)
(viii) \( 49x - 57y = 172 \); \( 57x - 49y = 252 \)
Answer: Let's solve each sub-part step-by-step.

(i) \( 3a + 5b = 26 \) ... (1)
\( a + 5b = 22 \) ... (2)
Subtracting equation (2) from (1):
\( (3a + 5b) - (a + 5b) = 26 - 22 \)
\( \implies 2a = 4 \)
\( \implies a = 2 \)
Substituting \( a = 2 \) in equation (2):
\( 2 + 5b = 22 \)
\( \implies 5b = 22 - 2 \)
\( \implies 5b = 20 \)
\( \implies b = 4 \)
The solution is \( (a, b) = (2, 4) \).

(ii) \( x + 7y = 10 \) ... (1)
\( 3x - 2y = 7 \) ... (2)
From equation (1), we get \( x = 10 - 7y \) ... (3)
Substituting this value of \( x \) in equation (2):
\( 3(10 - 7y) - 2y = 7 \)
\( \implies 30 - 21y - 2y = 7 \)
\( \implies 30 - 23y = 7 \)
\( \implies -23y = 7 - 30 \)
\( \implies -23y = -23 \)
\( \implies y = 1 \)
Substituting \( y = 1 \) in equation (3):
\( x = 10 - 7(1) \)
\( \implies x = 3 \)
The solution is \( (x, y) = (3, 1) \).

(iii) \( 2x - 3y = 9 \) ... (1)
\( 2x + y = 13 \) ... (2)
Subtracting equation (1) from (2):
\( (2x + y) - (2x - 3y) = 13 - 9 \)
\( \implies 4y = 4 \)
\( \implies y = 1 \)
Substituting \( y = 1 \) in equation (2):
\( 2x + 1 = 13 \)
\( \implies 2x = 12 \)
\( \implies x = 6 \)
The solution is \( (x, y) = (6, 1) \).

(iv) \( 5m - 3n = 19 \) ... (1)
\( m - 6n = -7 \) ... (2)
Multiply equation (2) by 5:
\( 5m - 30n = -35 \) ... (3)
Subtracting equation (3) from (1):
\( (5m - 3n) - (5m - 30n) = 19 - (-35) \)
\( \implies 27n = 54 \)
\( \implies n = 2 \)
Substituting \( n = 2 \) in equation (2):
\( m - 6(2) = -7 \)
\( \implies m - 12 = -7 \)
\( \implies m = 5 \)
The solution is \( (m, n) = (5, 2) \).

(v) \( 5x + 2y = -3 \) ... (1)
\( x + 5y = 4 \) ... (2)
Multiply equation (2) by 5:
\( 5x + 25y = 20 \) ... (3)
Subtracting equation (1) from (3):
\( (5x + 25y) - (5x + 2y) = 20 - (-3) \)
\( \implies 23y = 23 \)
\( \implies y = 1 \)
Substituting \( y = 1 \) in equation (2):
\( x + 5(1) = 4 \)
\( \implies x + 5 = 4 \)
\( \implies x = -1 \)
The solution is \( (x, y) = (-1, 1) \).

(vi) \( \frac{1}{3}x + y = \frac{10}{3} \) ... (1)
\( 2x + \frac{1}{4}y = \frac{11}{4} \) ... (2)
Multiply equation (1) by 3:
\( x + 3y = 10 \) ... (3)
Multiply equation (2) by 4:
\( 8x + y = 11 \) ... (4)
From equation (3), we get \( x = 10 - 3y \) ... (5)
Substituting this in equation (4):
\( 8(10 - 3y) + y = 11 \)
\( \implies 80 - 24y + y = 11 \)
\( \implies 80 - 23y = 11 \)
\( \implies -23y = 11 - 80 \)
\( \implies -23y = -69 \)
\( \implies y = 3 \)
Substituting \( y = 3 \) in equation (5):
\( x = 10 - 3(3) \)
\( \implies x = 1 \)
The solution is \( (x, y) = (1, 3) \).

(vii) \( 99x + 101y = 499 \) ... (1)
\( 101x + 99y = 501 \) ... (2)
Adding equations (1) and (2):
\( 200x + 200y = 1000 \)
Dividing by 200:
\( x + y = 5 \) ... (3)
Subtracting equation (1) from (2):
\( 2x - 2y = 2 \)
Dividing by 2:
\( x - y = 1 \) ... (4)
Adding equations (3) and (4):
\( 2x = 6 \)
\( \implies x = 3 \)
Substituting \( x = 3 \) in equation (3):
\( 3 + y = 5 \)
\( \implies y = 2 \)
The solution is \( (x, y) = (3, 2) \).

(viii) \( 49x - 57y = 172 \) ... (1)
\( 57x - 49y = 252 \) ... (2)
Adding equations (1) and (2):
\( 106x - 106y = 424 \)
Dividing by 106:
\( x - y = 4 \) ... (3)
Subtracting equation (1) from (2):
\( 8x + 8y = 80 \)
Dividing by 8:
\( x + y = 10 \) ... (4)
Adding equations (3) and (4):
\( 2x = 14 \)
\( \implies x = 7 \)
Substituting \( x = 7 \) in equation (4):
\( 7 + y = 10 \)
\( \implies y = 3 \)
The solution is \( (x, y) = (7, 3) \). These solutions demonstrate how different algebraic methods can be applied depending on the structure of the coefficients.
In simple words: We solve these pairs of equations by either substituting one variable into another or by adding/subtracting the equations to eliminate one variable. For equations with swapped coefficients, adding and subtracting them first makes them much simpler to solve.

🎯 Exam Tip: For equations with large, symmetric coefficients like in (vii) and (viii), always add and subtract the equations first to get two very simple equations in the form \( x + y = a \) and \( x - y = b \).

Question 2(i). Solve the following simultaneous equations:
\( 3a + 5b = 26 \)
\( a + 5b = 22 \)
Answer:
Subtracting the second equation from the first gives \( 2a = 4 \), which simplifies to \( a = 2 \). Substituting this back helps us find \( b \).
\( \implies b = \frac{20}{5} = 4 \)
\( \implies (a, b) = (2, 4) \) is the solution of the given simultaneous equations.
In simple words: By solving the equations, we find that the value of \( a \) is 2 and the value of \( b \) is 4.

🎯 Exam Tip: Double-check your final values by substituting them back into both original equations to make sure they satisfy both.

Question 2(ii). Solve the following simultaneous equations:
\( x + 7y = 10 \)
\( 3x - 2y = 7 \)
Answer:
\( x + 7y = 10 \)
\( \implies x = 10 - 7y \quad \text{--- (i)} \)
\( 3x - 2y = 7 \quad \text{--- (ii)} \)
Substituting \( x = 10 - 7y \) in equation (ii), we get:
\( 3(10 - 7y) - 2y = 7 \)
\( \implies 30 - 21y - 2y = 7 \)
\( \implies -23y = 7 - 30 \)
\( \implies -23y = -23 \)
\( \implies y = \frac{-23}{-23} \)
\( \implies y = 1 \)
Substituting \( y = 1 \) in equation (i), we get:
\( x = 10 - 7(1) \)
\( \implies x = 10 - 7 \)
\( \implies x = 3 \)
\( \implies (x, y) = (3, 1) \) is the solution of the given simultaneous equations.
In simple words: We find the value of \( x \) from the first equation and put it into the second equation to find \( y \). Once we get \( y = 1 \), we plug it back to find \( x = 3 \).

🎯 Exam Tip: Always write down the final solution as coordinates \( (x, y) \) clearly at the end to ensure you don't lose presentation marks.

Question 2(iii). Solve the following simultaneous equations:
\( 2x - 3y = 9 \)
\( 2x + y = 13 \)
Answer:
\( 2x - 3y = 9 \quad \text{--- (i)} \)
\( 2x + y = 13 \quad \text{--- (ii)} \)
Subtracting equation (ii) from (i), we get:

\( \implies -4y = -4 \)
\( \implies y = \frac{-4}{-4} \)
\( \implies y = 1 \)
Substituting \( y = 1 \) in equation (ii), we get:
\( 2x + 1 = 13 \)
\( \implies 2x = 12 \)
\( \implies x = \frac{12}{2} \)
\( \implies x = 6 \)
\( \implies (x, y) = (6, 1) \) is the solution of the given simultaneous equations.
In simple words: Since both equations have \( 2x \), we subtract one equation from the other to eliminate \( x \) and solve for \( y \). Then we use \( y \) to find \( x \).

🎯 Exam Tip: When subtracting equations, remember to change the signs of all terms in the second equation to avoid calculation errors.

Question 2(iv). Solve the following simultaneous equations:
\( 5m - 3n = 19 \)
\( m - 6n = -7 \)
Answer:
\( 5m - 3n = 19 \quad \text{--- (i)} \)
\( m - 6n = -7 \)
\( \implies m = 6n - 7 \quad \text{--- (ii)} \)
Substituting \( m = 6n - 7 \) in equation (i), we get:
\( 5(6n - 7) - 3n = 19 \)
\( \implies 30n - 35 - 3n = 19 \)
\( \implies 27n = 19 + 35 \)
\( \implies 27n = 54 \)
\( \implies n = \frac{54}{27} \)
\( \implies n = 2 \)
Substituting \( n = 2 \) in equation (ii), we get:
\( m = 6(2) - 7 \)
\( \implies m = 12 - 7 \)
\( \implies m = 5 \)
\( \implies (m, n) = (5, 2) \) is the solution of the given simultaneous equations.
In simple words: We express \( m \) in terms of \( n \) from the second equation, substitute it into the first equation to find \( n \), and then calculate \( m \).

🎯 Exam Tip: Expressing a variable with a coefficient of 1 (like \( m \) here) in terms of the other variable is often the easiest way to use the substitution method.

Question 2. (iv) Solve the following simultaneous equations:
\( 5m - 3n = 19 \); \( m - 6n = -7 \)
Answer:
\( 5m - 3n = 19 \) ...(i)
\( m - 6n = -7 \)
\( \therefore m = 6n - 7 \) ...(ii)
Substituting this value of \( m \) in equation (i), we get:
\( 5(6n - 7) - 3n = 19 \)
\( \therefore 30n - 35 - 3n = 19 \)
\( \therefore 27n = 19 + 35 \)
\( \dots \)
\( \therefore 27n = 54 \)
\( \therefore n = \frac{54}{27} = 2 \)
Substituting \( n = 2 \) in equation (ii), we get
\( m = 6(2) - 7 \)
\( = 12 - 7 = 5 \)
\( \dots \)
\( \therefore (m, n) = (5, 2) \) is the solution of the given simultaneous equations.
In simple words: We find the values of \( m \) and \( n \) by substituting one equation into the other, giving us the final answer of \( m = 5 \) and \( n = 2 \).

🎯 Exam Tip: Always write the final answer in the coordinate format \( (m, n) \) and double-check by substituting the values back into the original equations.

Question 2. (v) Solve the following simultaneous equations:
\( 5x + 2y = -3 \); \( x + 5y = 4 \)
Answer:
\( 5x + 2y = -3 \) ...(i)
\( x + 5y = 4 \)
\( \therefore x = 4 - 5y \) ...(ii)
Substituting \( x = 4 - 5y \) in equation (i), we get
\( 5(4 - 5y) + 2y = -3 \)
\( \therefore 20 - 25y + 2y = -3 \)
\( \therefore -23y = -3 - 20 \)
\( \therefore -23y = -23 \)
\( \therefore y = \frac{-23}{-23} = 1 \)
Substituting \( y = 1 \) in equation (ii), we get
\( x = 4 - 5(1) \)
\( = 4 - 5 = -1 \)
\( \therefore (x, y) = (-1, 1) \) is the solution of the given simultaneous equations.
In simple words: We express \( x \) in terms of \( y \) from the second equation, substitute it into the first equation to find \( y = 1 \), and then use that to find \( x = -1 \).

🎯 Exam Tip: Be very careful with negative signs when dividing, such as \( \frac{-23}{-23} = 1 \), as a simple sign error can throw off the entire solution.

Question 2(vi). Solve the simultaneous equations: \( \frac{1}{3}x + y = \frac{10}{3} \); \( 2x + \frac{1}{4}y = \frac{11}{4} \)
Answer:
\( \frac{1}{3}x + y = \frac{10}{3} \)
\( \implies x + 3y = 10 \) ...[Multiplying both sides by 3]
\( \implies x = 10 - 3y \) ...(i)
\( 2x + \frac{1}{4}y = \frac{11}{4} \)
\( \implies 8x + y = 11 \) ...(ii) [Multiplying both sides by 4]
Substituting \( x = 10 - 3y \) in equation (ii), we get:
\( 8(10 - 3y) + y = 11 \)
\( \implies 80 - 24y + y = 11 \)
\( \implies -23y = 11 - 80 \)
\( \implies -23y = -69 \)
\( \implies y = \frac{-69}{-23} = 3 \)
Substituting \( y = 3 \) in equation (i), we get:
\( x = 10 - 3(3) \)
\( = 10 - 9 = 1 \)
\( \implies (x, y) = (1, 3) \) is the solution of the given simultaneous equations.
In simple words: To solve equations with fractions, we first multiply them by the denominators to clear the fractions. Then we find the value of one variable and substitute it to find the other.

🎯 Exam Tip: Always multiply the entire equation by the denominator to eliminate fractions first, which makes calculations much simpler and reduces the chance of errors.

Question 2(vii). Solve the simultaneous equations: \( 99x + 101y = 499 \); \( 101x + 99y = 501 \)
Answer:
\( 99x + 101y = 499 \) ...(i)
\( 101x + 99y = 501 \) ...(ii)
Adding equations (i) and (ii), we get:
\( 200x + 200y = 1000 \)
Dividing both sides by 200, we get:
\( \implies x + y = 5 \) ...(iii)
Subtracting equation (i) from (ii), we get:
\( 2x - 2y = 2 \)
Dividing both sides by 2, we get:
\( \implies x - y = 1 \) ...(iv)
Adding equations (iii) and (iv), we get:
\( 2x = 6 \)
\( \implies x = 3 \)
Substituting \( x = 3 \) in equation (iii), we get:
\( 3 + y = 5 \)
\( \implies y = 2 \)
\( \implies (x, y) = (3, 2) \) is the solution of the given simultaneous equations.
In simple words: When the numbers in front of x and y are swapped between the two equations, we can add and subtract the equations to get much simpler equations to solve.

🎯 Exam Tip: When you see large, swapped coefficients like 99 and 101, do not try to multiply them directly. Always add and subtract the equations first to simplify them.

Question 2. (vii) Solve the following simultaneous equations: \( 99x + 101y = 499 \); \( 101x + 99y = 501 \)
Answer:
\( 99x + 101y = 499 \) ...(i)
\( 101x + 99y = 501 \) ...(ii)

Adding equations (i) and (ii), we get:
\( \quad 99x + 101y = 499 \)
\( + 101x + 99y = 501 \)
-------------------------
\( \quad 200x + 200y = 1000 \)

Dividing both sides by 200:
\( \quad \frac{200x}{200} + \frac{200y}{200} = \frac{1000}{200} \)

\( \implies x + y = 5 \) ...(iii)

Subtracting equation (ii) from (i), we get:
\( \quad 99x + 101y = 499 \)
\( - (101x + 99y = 501) \)
-------------------------
\( \quad -2x + 2y = -2 \)

Dividing both sides by -2:
\( \quad \frac{-2x}{-2} + \frac{2y}{-2} = \frac{-2}{-2} \)

\( \implies x - y = 1 \) ...(iv)

Adding equations (iii) and (iv), we get:
\( \quad x + y = 5 \)
\( + x - y = 1 \)
-------------------------
\( \quad 2x = 6 \)

\( \implies x = \frac{6}{2} \)

\( \implies x = 3 \)

Substituting \( x = 3 \) in equation (iii), we get:
\( 3 + y = 5 \)

\( \implies y = 5 - 3 \)

\( \implies y = 2 \)

\( \therefore (x, y) = (3, 2) \) is the solution of the given simultaneous equations.
In simple words: When the numbers in front of x and y are swapped between the two equations, we can add them once and subtract them once to get two very simple equations. Solving those simple equations gives us the final answer easily.

🎯 Exam Tip: When you see large, interchanged coefficients like 99 and 101, do not try to multiply them directly. Always use the "Add-Subtract-Add" method to save time and avoid calculation errors.

Question 2. (viii) Solve the following simultaneous equations: \( 49x - 57y = 172 \); \( 57x - 49y = 252 \)
Answer:
\( 49x - 57y = 172 \) ...(i)
\( 57x - 49y = 252 \) ...(ii)

Adding equations (i) and (ii), we get:
\( \quad 49x - 57y = 172 \)
\( + 57x - 49y = 252 \)
-------------------------
\( \quad 106x - 106y = 424 \)

Dividing both sides by 106:
\( \quad \frac{106x}{106} - \frac{106y}{106} = \frac{424}{106} \)

\( \implies x - y = 4 \) ...(iii)

Subtracting equation (i) from (ii), we get:
\( \quad 57x - 49y = 252 \)
\( - (49x - 57y = 172) \)
-------------------------
\( \quad 8x + 8y = 80 \)

Dividing both sides by 8:
\( \quad \frac{8x}{8} + \frac{8y}{8} = \frac{80}{8} \)

\( \implies x + y = 10 \) ...(iv)

Adding equations (iii) and (iv), we get:
\( \quad x - y = 4 \)
\( + x + y = 10 \)
-------------------------
\( \quad 2x = 14 \)

\( \implies x = \frac{14}{2} \)

\( \implies x = 7 \)

Substituting \( x = 7 \) in equation (iv), we get:
\( 7 + y = 10 \)

\( \implies y = 10 - 7 \)

\( \implies y = 3 \)

\( \therefore (x, y) = (7, 3) \) is the solution of the given simultaneous equations.
In simple words: Just like the previous problem, we add and subtract the equations to simplify them into two basic equations, which we then solve to find the values of x and y.

🎯 Exam Tip: Be extremely careful with signs when subtracting equations. A negative sign outside the bracket changes the signs of all terms inside, which is a very common place to make mistakes.

Question. Solve the simultaneous equations: \( 49x - 57y = 172 \) and \( 57x - 49y = 252 \)
Answer: Let the given equations be:
\( 49x - 57y = 172 \) ---(i)
\( 57x - 49y = 252 \) ---(ii)
Adding equations (i) and (ii), we get: \[ \begin{array}{rcc} 49x - 57y & = & 172 \\ +\quad 57x - 49y & = & 252 \\ \hline 106x - 106y & = & 424 \end{array} \]
\( \implies x - y = \frac{424}{106} \) ...[Dividing both sides by 106]
\( \implies x - y = 4 \) ---(iii)
Subtracting equation (ii) from (i), we get: \[ \begin{array}{rcc} 49x - 57y & = & 172 \\ -\quad 57x - 49y & = & 252 \\ (\text{--}) \quad (+) & & (\text{--}) \\ \hline -8x - 8y & = & -80 \end{array} \]
\( \implies x + y = \frac{-80}{-8} \) ...[Dividing both sides by -8]
\( \implies x + y = 10 \) ---(iv)
Adding equations (iii) and (iv), we get: \[ \begin{array}{rcc} x - y & = & 4 \\ +\quad x + y & = & 10 \\ \hline 2x & = & 14 \end{array} \]
\( \implies x = \frac{14}{2} = 7 \)
Substituting \( x = 7 \) in equation (iv), we get:
\( 7 + y = 10 \)
\( \implies y = 10 - 7 = 3 \)
\( \implies (x, y) = (7, 3) \) is the solution of the given simultaneous equations. This systematic approach of adding and subtracting equations is particularly useful when the coefficients of the variables are large and symmetrical.
In simple words: When we have equations where the coefficients of x and y are interchanged, we can solve them easily by adding and subtracting the equations to get two simpler equations.

🎯 Exam Tip: When coefficients of \( x \) and \( y \) are interchanged, always use the 'Add-Subtract-Add' (ASA) method to find the values of the variables quickly and avoid calculation errors.

Question. Complete the following table. (Textbook pg. no. 1)
Answer:

Identifying whether an equation is linear helps us determine if its graph will be a straight line.
In simple words: A linear equation in two variables must have exactly two variables, and the highest power (degree) of each variable must be 1.

🎯 Exam Tip: Remember that terms like \( \frac{1}{x} \) have a degree of -1, and terms like \( xy \) have a degree of 2, so they cannot be part of a linear equation in two variables.

Question 1. Solve: \( 3x + 2y = 29 \); \( 5x - y = 18 \) (Textbook pg. no. 3)
Answer: Let the given equations be:
\( 3x + 2y = 29 \) ---(i)
\( 5x - y = 18 \) ---(ii)
Multiplying equation (ii) by 2, we get:
\( 10x - 2y = 36 \) ---(iii)
Adding equations (i) and (iii), we get: \[ \begin{array}{rcc} 3x + 2y & = & 29 \\ +\quad 10x - 2y & = & 36 \\ \hline 13x & = & 65 \end{array} \]
\( \implies x = \frac{65}{13} \)
\( \implies x = 5 \)
Substituting \( x = 5 \) in equation (ii), we get:
\( 5(5) - y = 18 \)
\( \implies 25 - y = 18 \)
\( \implies y = 25 - 18 \)
\( \implies y = 7 \)
\( \therefore (x, y) = (5, 7) \) is the solution of the given simultaneous equations. Eliminating one variable by equating coefficients is a highly reliable method for solving simultaneous linear equations.
In simple words: We can solve these equations by making the coefficients of \( y \) equal but opposite in sign, then adding them to eliminate \( y \) and find \( x \).

🎯 Exam Tip: Always double-check your final values of \( x \) and \( y \) by substituting them back into one of the original equations to ensure they satisfy it.

Question 1. Let's solve the equations by eliminating 'y'. Fill suitably the boxes below.
\( 3x + 2y = 29 \) ...(i)
and \( 5x - y = 18 \) ...(ii)
Answer:
Multiplying equation (ii) by 2, we get
\( 5x \times \boxed{2} - y \times \boxed{2} = 18 \times \boxed{2} \)
\( \therefore 10x - 2y = \boxed{36} \) ...(iii)
Add equations (i) and (iii).
\( \begin{array}{r@{\quad}l} 3x + 2y &= 29 \\ + \quad \boxed{10x} - \boxed{2y} &= \boxed{36} \\ \hline \boxed{13x} &= \boxed{65} \end{array} \)
\( \therefore x = \frac{65}{13} = \boxed{5} \)
Substituting \( x = 5 \) in equation (i).
\( 3x + 2y = 29 \)
\( \therefore 3 \times \boxed{5} + 2y = 29 \)
\( \dots \)
\( \therefore \boxed{15} + 2y = 29 \)
\( \therefore 2y = 29 - \boxed{15} \)
\( \dots \)
\( \therefore 2y = \boxed{14} \)
\( \therefore y = \frac{14}{2} = \boxed{7} \)
\( \therefore (x, y) = (\boxed{5}, \boxed{7}) \) is the solution.
In simple words: To solve these equations, we multiply the second equation by 2 so that the 'y' terms cancel out when added. This lets us find 'x' first, which we then use to find 'y'.

🎯 Exam Tip: Make sure to fill every single box carefully, as each box carries specific marks in activity-based questions.