Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 5 Linear Equations in Two Variables Set 5.3 Solutions

Question 1. Choose the correct alternative answers for the following questions.
i. If 3x + 5y = 9 and 5x + 3y = 7, then what is the value of x + y ?
(A) 2
(B) 16
(C) 9
(D) 7
Answer: (A) 2
ii. 'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26'. What is the mathematical form of the statement ?
(A) x - y = 8
(B) x + y = 8
(C) x + y = 23
(D) 2x + y = 21
Answer: (C) x + y = 23
iii. Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's age?
(A) 20 years
(B) 15 years
(C) 10 years
(D) 5 years
Answer: (C) 10 years
Hints:
i. Adding the given equations,
3x+ 5y = 9
5x + 3y = 7
8x + 8y = 16
.
. x + y = 2 . [Dividing both sides by 8]
ii. Let the length of the rectangle be 'x' and that of breadth be 'y'.
Perimeter of rectangle = 2[(x - 5) + (y - 5)]
.
. 26 = 2(x + y - 10)
.
. x + y - 10 = 13
∴ x + y = 23
iii. Let the age of Ajay be x years.
.
. x + (x + 5) = 25
.
. 2x = 20
.
. x = 10 years
In simple words: This question tests your ability to form and solve linear equations based on given conditions, especially for scenarios like age, perimeter, and algebraic sums. It involves combining equations or setting up expressions to find unknown variables.

🎯 Exam Tip: Pay close attention to keywords like "sum," "difference," "perimeter," and "interchanging digits" as they directly translate into mathematical operations and relationships. Double-check your calculations, especially when dealing with signs and combining terms.

Question 2. Solve the following simultaneous equations.
i. 2x + y = 5 ; 3x - y = 5
ii. x - 2y = -1 ; 2x - y = 7
iii. x + y = 11 ; 2x - 3y = 7
iv. 2x + y = -2 ; 3x - y = 7
V. 2x - y = 5; 3x + 2y = 11
vi. x - 2y - 2 ; x + 2y = 10
Answer:
i. 2x + y = 5 ...(i)
3x - y = 5 ...(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x - y = 5
5x = 10
.
. x = 10/5
.
. x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 - 4 = 1
.
. (2, 1) is the solution of the given equations.
ii. x - 2y = -1
.
.x = 2y - 1 ....(i)
.
. 2x - y = 7 ....(ii)
Substituting x = 2y - 1 in equation (ii),
2(2y - 1) - y = 7
.
. 4y - 2 - y = 7
.
. 3y = 7 + 2
.
. 3y = 9
∴ y = 9/3
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y - 1
.
. x = 2(3) - 1
∴ x = 6 - 1 = 5
.
. (5, 3) is the solution of the given equations.
iii. x + y = 11
.
. x = 11 - y ...(i)
2x - 3y = 7 .......(ii)
Substituting x = 11 -y in equation (ii),
2(11 - y) - 3y = 7
.
. 22 - 2y - 3y = 7
.
. 22 - 5y = 7
.
. 22 - 7 = 5y
.
. 15 = 5y
\( \frac{15}{5} \)
∴ y = 3
Substituting y = 3 in equation (i),
x = 11-y
.
. x = 11-3 = 8
.
. (8, 3) is the solution of the given equations.
iv. 2x + y = -2 ...(i)
3x - y = 7 ...(ii)
Adding equations (i) and (ii),
2x + y = -2
+ 3x - y = 7
5x = 5
\( \frac{5}{5} \)
∴ x = 1
Substituting x = 1 in equation (i),
2x + y = -2
.
. 2(1) +y = -2
2 + y = -2
∴ y = - 2 - 2
∴ y = -4
.
. (1, -4) is the solution of the given equations.
v. 2x - y = 5
.: -y = 5 - 2x
∴ y = 2x - 5 ...(i)
3x + 2y = 11 ......(ii)
Substituting y = 2x - 5 in equation (ii),
3x + 2(2x - 5) = 11
.
. 3x + 4x- 10= 11
.
. 7x = 11 + 10
.
. 7x = 21
\( \frac{21}{7} \)
∴ x = 3
Substituting x = 3 in equation (i),
y = 2x - 5
∴ y = 2(3) - 5
∴ y = 6 - 5 = 1
.(3,1) is the solution of the given equations.
vi. x - 2y = -2
.
. x = 2y - 2 ...(i)
x + 2y = 10 .....(ii)
Substituting x = 2y - 2 in equation (ii),
2y - 2 + 2y = 10
.
. 4y = 10 + 2
.
. 4y= 12
\( \frac{12}{4} \)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y - 2
.
. x = 2(3) - 2
.
. x = 6 - 2 = 4
.
. (4, 3) is the solution of the given equations.
In simple words: This question requires you to solve various pairs of simultaneous linear equations using methods like substitution or elimination. The goal is to find the unique values for 'x' and 'y' that satisfy both equations in each pair.

🎯 Exam Tip: When solving simultaneous equations, choose the method (substitution or elimination) that seems easiest for the given equations. Always verify your solution by substituting the found 'x' and 'y' values back into both original equations to ensure correctness.

Question 3. By equating coefficients of variables, solve the following equations.
i. 3x - 4y = 7; 5x + 2y = 3
ii. 5x + 7y= 17 ; 3x - 2y = 4
iii. x - 2y = -10 ; 3x - 5y = -12
iv. 4x+y = 34 ; x + 4y = 16
Answer:
i. 3x - 4y = 7 ...(i)
5x + 2y = 3 ....(ii)
Multiplying equation (ii) by 2,
10x + 4y = 6 ...(iii)
Adding equations (i) and (iii),
3x - 4y = 7
+ 10x + 4y = 6
––––––––––––––
13x = 13
.
. \( x = \frac{13}{13} \)
.
. x = 1
Substituting x = 1 in equation (i),
3x - 4y = 7
.: 3(1) - 4y = 7
.
. 3 - 4y = 7
.
. 3 - 7 = 4y
.
. -4 = 4y
: \( y = \frac{-4}{4} \)
∴ y = -1
.
. (1, -1) is the solution of the given equations.
ii. 5x + 7y = 17 ...(i)
3x - 2y = 4 ....(ii)
Multiplying equation (i) by 2,
10x + 14y = 34 ...(iii)
Multiplying equation (ii) by 7,
21x - 14y = 28 .....(iv)
Adding equations (iii) and (iv),
10x + 14y = 34
+ 21x - 14y = 28
––––––––––––––
31x = 62
.
. \( x = \frac{62}{31} \)
.: x = 2
Substituting x = 2 in equation (ii),
3x - 2y = 4
.
. 3(2) - 2y = 4
.
. 6 - 2y = 4
.
. 6 - 4 = 2y
.
. 2 = 2y
\( \frac{2}{2} \)
∴ y = 1
.
. (2,1) is the solution of the given equations.
iii. x - 2y = -10 ....(i)
3x - 5y = -12 .......(ii)
Multiplying equation (i) by 3,
3x - 6y = -30 ...(iii)
Subtracting equation (ii) from (iii),
3x - 6y = -30
- (3x - 5y = -12)
––––––––––––––
-y = -18
∴ y = 18
Substituting y = 18 in equation (i),
x - 2y = -10
.
. x - 2(18) = -10
.
. x - 36 = -10
.
. x = -10 + 36 = 26
.
. (26, 18) is the solution of the given equations.
iv. 4x + y = 34 ...(i)
x + 4y = 16 ...... (ii)
Multiplying equation (i) by 4,
16x + 4y = 136 ...(iii)
Subtracting equation (ii) from (iii),
16x + 4y = 136
- (x + 4y = 16)
––––––––––––––
15x = 120
.
. \( x = \frac{120}{15} \)
x = 8
Substituting x = 8 in equation (i),
4x + y = 34
.
. 4(8) + y = 34
.
. 32 + y = 34
∴ y = 34 - 32 = 2
.
. (8, 2) is the solution of the given equations.
In simple words: This question demonstrates how to solve simultaneous linear equations by making the coefficients of one variable equal in both equations, then adding or subtracting the equations to eliminate that variable and solve for the other.

🎯 Exam Tip: The key to the elimination method is finding the least common multiple of the coefficients you wish to eliminate. Remember to multiply the *entire* equation (both sides) by the chosen factor to maintain equality. Always double-check your arithmetic, especially when dealing with negative numbers.

Question 4. Solve the following simultaneous equations.
i. \( \frac{x}{3} + \frac{y}{4} = 4; \frac{x}{2} - \frac{y}{4} = 1 \)
ii. \( \frac{x}{3} + 5y = 13; 2x + \frac{y}{3} = 19 \)
iii. \( \frac{2}{x} + \frac{3}{y} = 13; \frac{5}{x} - \frac{4}{y} = -2 \)
Answer:
i. \( \frac{x}{3} + \frac{y}{4} = 4 \)
Multiplying both sides by 12,
4x + 3y = 48 ...(i)
\( \frac{x}{2} - \frac{y}{4} = 1 \)
Multiplying both sides by 8,
4x - 2y = 8 .....(ii)
Subtracting equation (ii) from (i),
4x + 3y = 48
- (4x - 2y = 8)
––––––––––––––
5y = 40
.
. \( y = \frac{40}{5} \)
∴ y = 8
Substituting y = 8 in equation (ii),
4x - 2y = 8
.
. 4x - 2(8) = 8
.
. 4x - 16 = 8
.
. 4x = 8+ 16
.
. 4x = 24
.
. \( x = \frac{24}{4} \)
.
. x = 6
.
. (6, 8) is the solution of the given equations.
ii. \( \frac{x}{3} + 5y = 13 \)
Multiplying both sides by 3,
x + 15y = 39 ...(i)
\( 2x + \frac{y}{3} = 19 \)
Multiplying both sides by 2,
4x + y = 38 .......(ii)
Multiplying equation (i) by 4,
4x + 60y = 156 ...(iii)
Subtracting equation (ii) from (iii),
4x + 60y = 156
- (4x + y = 38)
––––––––––––––
59y = 118
.
. \( y = \frac{118}{59} \)
∴ y = 2
Substituting y = 2 in equation (i),
x + 15y = 39
.: x+ 15(2) = 39
.
. x + 30 = 39
.
. x = 39 - 30 = 9
.
. (9,2) is the solution of the given equations.
iii. \( \frac{2}{x} + \frac{3}{y} = 13 \)
Multiplying both sides by 5,
\( \frac{10}{x} + \frac{15}{y} = 65 \)
.
. \( \frac{10}{x} = 65 - \frac{15}{y} \)..(i)
\( \frac{5}{x} - \frac{4}{y} = -2 \)
Multiplying both sides by 2,
\( \frac{10}{x} - \frac{8}{y} = -4 \)
.
. \( \frac{10}{x} = \frac{8}{y} - 4 \)..(ii)
.
. \( 65 - \frac{15}{y} = \frac{8}{y} - 4 \)..[From (i) and (ii)]
.: \( 65 + 4 = \frac{8}{y} + \frac{15}{y} \)
.
. \( 69 = \frac{23}{y} \)
.
. \( y = \frac{23}{69} \)
\( y = \frac{1}{3} \)
Substituting \( y = \frac{1}{3} \) in equation (ii),
\( \frac{10}{x} - \frac{8}{y} = -4 \)
.
. \( \frac{10}{x} - \frac{8}{1/3} = -4 \)
.
. \( \frac{10}{x} = 8 \times 3 - 4 \)
.
. \( \frac{10}{x} = 24 - 4 \)
.
. \( \frac{10}{x} = 20 \)
.
. \( \frac{10}{20} = x \)
.
. \( x = \frac{1}{2} \)
.
. \( (\frac{1}{2}, \frac{1}{3}) \) is the solution of the given equations.
In simple words: This question involves solving simultaneous equations where variables might appear in the denominator or as fractions. The key is to clear the denominators by multiplying by appropriate factors or to make a substitution to simplify the equations into standard linear form.

🎯 Exam Tip: When equations involve fractions, always simplify them into standard linear form \(Ax + By = C\) first by multiplying by the least common multiple of the denominators. For equations with variables in the denominator (like \( \frac{1}{x} \)), consider substituting \(m = \frac{1}{x}\) and \(n = \frac{1}{y}\) to transform them into linear equations in terms of 'm' and 'n'.

Question 5. A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Answer:
Let the digit in unit's place be 'x' and the digit in ten's place be 'y'.

According to the first condition,
a two digit number is 3 more than 4 times the sum of its digits.
10y + x = 4(x + y) + 3
.
. 10y + x = 4x + 4y + 3
.
. x - 4x + 10y - 4y = 3
.: - 3x + 6y = 3
Dividing both sides by -3,
x - 2y = -1 ...(i)
According to the second condition,
if 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18= 10x + y
.
. x - 10x + 10y - y = -18
.
. - 9x + 9y = -18
Dividing both sides by - 9,
x - y = 2 ......(ii)
Subtracting equation (ii) from (i),
x - 2y = -1
- (x - y = 2)
––––––––––––––
-y = -3
∴ y = 3
Substituting y = 3 in equation (ii),
x - y = 2
∴ x - 3 = 2
∴ x = 2 + 3 = 5
.
. Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
The required number is 35.
In simple words: This problem involves setting up two linear equations based on conditions given for a two-digit number and its digits. One equation relates the number to the sum of its digits, and the other involves interchanging the digits. Solving these equations reveals the original number.

🎯 Exam Tip: For problems involving two-digit numbers, always define the tens digit as 'y' and units digit as 'x'. The number itself will then be \(10y + x\). When digits are interchanged, the new number is \(10x + y\). Carefully translate the word problem into mathematical equations, paying attention to positive and negative signs.

Question 6. The total cost of 8 books and 5 pens is Rs. 420 and the total cost of 5 books and 8 pens is Rs. 321. Find the cost of 1 book and 2 pens.
Answer:
Let the cost of one book be x and the cost of one pen be y.
According to the first condition,
the total cost of 8 books and 5 pens is Rs. 420.
.
. 8x + 5y = 420 ...(i)
According to the second condition, the total cost of 5 books and 8 pens is Rs. 321.
5x + 8y = 321 ....(ii)
Multiplying equation (i) by 5,
40x + 25y = 2100 ...(iii)
Multiplying equation (ii) by 8,
40x + 64y = 2568 ... (iv)
Subtracting equation (iii) from (iv),
40x + 64y = 2568
- (40x + 25y = 2100)
––––––––––––––
39y = 468
.
. \( y = \frac{468}{39} \)
∴ y = 12
Substituting y = 12 in equation (i),
8x + 5y = 420
.
. 8x + 5(12) = 420
.
. 8x + 60 = 420
.
. 8x = 420 - 60
.
. 8x = 360
.
. \( x = \frac{360}{8} \)
.
. x = 45
Cost of 1 book and 2 pens = x + 2y
= 45 + 2(12)
= 45 + 24
= Rs. 69
.
. The cost of 1 book and 2 pens is Rs. 69.
In simple words: This problem asks you to find the cost of a book and a pen by setting up two linear equations based on the total cost of different quantities of books and pens, then solving them simultaneously. Finally, calculate the cost of 1 book and 2 pens.

🎯 Exam Tip: Clearly define your variables (e.g., 'x' for book cost, 'y' for pen cost) at the start. When solving, if one variable has large coefficients, consider eliminating the other variable first to simplify calculations. Remember to answer the specific question asked, which in this case is the cost of 1 book and 2 pens, not just 'x' and 'y'.

Question 7. The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves Rs. 200, find the income of each.
Answer:
Let the income of first person be x and that of second person be y.
According to the first condition,
the ratio of their incomes is 9: 7.
.
. \( \frac{x}{y} = \frac{9}{7} \)
.
. 7x = 9y
.
. 7x - 9y = 0 .......(i)
Each person saves Rs. 200.
Expenses of first person = Income - Saving = x - 200
Expenses of second person = y - 200
According to the second condition,
the ratio of their expenses is 4 : 3
.
. \( \frac{x-200}{y-200} = \frac{4}{3} \)
.: 3(x - 200) = 4(y - 200)
.
. 3x - 600 = 4y - 800
.
. 3x - 4y = -800 + 600
.
. 3x - 4y = -200 ...(ii)
Multiplying equation (i) by 4,
28x - 36y = 0 ...(iii)
Multiplying equation (ii) by 9,
27x - 36y = -1800 ...(iv)
Subtracting equation (iv) from (iii),
28x - 36y = 0
- (27x - 36y = -1800)
––––––––––––––
x = 1800
Substituting x = 1800 in equation (i),
7x - 9y = 0
.
. 7(1800) - 9y = 0
.
. 9y = 7 x 1800
∴ \( y = \frac{7 \times 1800}{9} \)
y = 7 x 200
.: y = 1400
.
. The income of first person is Rs. 1800 and that of second person is Rs. 1400.
In simple words: This problem involves using ratios to set up linear equations. You define incomes and expenses using variables and the given savings. Two equations are formed from the income ratio and expense ratio, which are then solved to find individual incomes.

🎯 Exam Tip: When dealing with ratios, represent the actual quantities using a common multiple (e.g., if ratio is A:B, then A=k*ratioA, B=k*ratioB). Remember the formula: Income - Savings = Expenses. Be meticulous with cross-multiplication and algebraic manipulations to avoid errors.

Question 8. If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 9 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Answer:
Let the length of the rectangle be 'x' units and the breadth of the rectangle be 'y' units.
Area of the rectangle = xy sq. units
length of the rectangle is reduced by 5 units
.
. length = x -5
breadth of the rectangle is increased by 3 units
.
. breadth = y + 3
area of the rectangle is reduced by 9 square units
.
. area of the rectangle = xy - 9
According to the first condition,
(x - 5)(y + 3) = xy - 9
.
. xy + 3x - 5y - 15 = xy - 9
.
. 3x - 5y = -9 + 15
.
. 3x - 5y = 6 ...(i)
length of the rectangle is reduced by 3 units
.
. length = x - 3
breadth of the rectangle is increased by 2 units
.
. breadth = y + 2
area of the rectangle is increased by 67 square units
.
. area of the rectangle = xy + 61
According to the second condition,
(x - 3)(y + 2) = xy + 67
.
. xy + 2x - 3y - 6 = xy + 67
.
. 2x - 3y = 67 + 6
.
. 2x - 3y = 73 ...(ii)
Multiplying equation (i) by 3,
9x - 15y = 18 ...(iii)
Multiplying equation (ii) by 5,
10x - 15y = 365 ...(iv)
Subtracting equation (iii) from (iv),
10x - 15y = 365
- (9x - 15y = 18)
––––––––––––––
x = 347
Substituting x = 347 in equation (ii),
2x - 3y = 73
.
. 2(347) - 3y = 73
.: 694 - 73 = 3y
.
. 621 = 3y
∴ \( y = \frac{621}{3} \)
∴ y = 207
.
. The length and breadth of rectangle are 347 units and 207 units respectively.
In simple words: This problem requires you to form two linear equations by analyzing changes in the length, breadth, and area of a rectangle. You set up equations from two different scenarios, then solve them simultaneously to find the original length and breadth.

🎯 Exam Tip: Define initial length and breadth as 'x' and 'y', respectively, and the original area as 'xy'. Carefully translate each condition into an equation involving the modified dimensions and areas. Remember to expand products like \((x-5)(y+3)\) correctly using the distributive property, and simplify to standard linear forms before solving.

Subtracting equation (iii) from (iv), \( 10x - 15y = 365 \) \( 9x - 15y = 18 \)
\( \underline{ - \quad + \quad -} \)
\( \underline{ x \quad \quad = 347 } \)
Substituting \( x = 347 \) in equation (ii),
\( 2x - 3y = 73 \)
\( \therefore 2(347) - 3y = 73 \)
\( \therefore 694 - 73 = 3y \)
\( \therefore 621 = 3y \)
\( \therefore y = \frac{621}{3} \)
\( \therefore y = 207 \)
\( \therefore \) The length and breadth of rectangle are 347 units and 207 units respectively.
In simple words: By setting up simultaneous equations based on the given conditions about changes in rectangle dimensions and area, we solved for the length and breadth, finding them to be 347 units and 207 units respectively.

🎯 Exam Tip: When solving word problems involving two-variable linear equations, carefully translate each condition into an algebraic equation. Pay attention to signs during substitution and elimination.

Question 9. The distance between two places A and B on a road is 70 kilometres. A car starts from A and the other from B. If they travel in the same direction, they will meet in 7 hours. If they travel towards each other they will meet in 1 hour, then find their speeds.
Answer:
Let the speed of the car starting from A (first car) be 'x' km/hr and that starting from B (second car) be 'y' km/hr. (x > y)
According to the first condition,
Distance covered by the first car in 7 hours = 7x km
Distance covered by the second car in 7 hours = 7y km
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो कारों की गति की स्थिति को दर्शाता है। बिंदु A और B के बीच की दूरी 70 किमी है। पहली कार A से 7x किमी की दूरी तय करती है, जबकि दूसरी कार B से 7y किमी की दूरी तय करती है, जब दोनों एक ही दिशा में यात्रा करती हैं।
If the cars are travelling in the same direction, \( 7x - 7y = 70 \)
Dividing both sides by 7,
\( x - y = 10 \)...(i)
According to the second condition,
Distance covered by the first car in
1 hour = x km
Distance covered by the second car in 1 hour = y km
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विपरीत दिशाओं में यात्रा कर रही दो कारों की स्थिति को दिखाता है। बिंदु A और B के बीच की कुल दूरी 70 किमी है। कार A से x किमी की दूरी तय करती है, जबकि कार B से y किमी की दूरी तय करती है, जब वे एक-दूसरे की ओर यात्रा करती हैं।
If the cars are travelling in the opposite direction
\( x + y = 70 \)...(ii)
Adding equations (i) and (ii),
\( x - y = 10 \)
\( \underline{ + x + y = 70 } \)
\( \underline{ 2x \quad = 80 } \)
\( \therefore x = \frac{80}{2} \)
\( \therefore x = 40 \)
Substituting \( x = 40 \) in equation (ii), \( x + y = 70 \)
\( \therefore 40 + y = 70 \)
\( \therefore y = 70 - 40 = 30 \)
\( \therefore \) The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.
In simple words: We set up two equations based on the cars' relative speeds and distances covered when traveling in the same and opposite directions. Solving these simultaneous equations, we found the speeds of the two cars to be 40 km/hr and 30 km/hr.

🎯 Exam Tip: For problems involving relative motion, remember that speeds add up when moving towards each other and subtract when moving in the same direction. Clear diagrammatic representation helps in visualizing the problem.

Question 10. The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Answer:
Let the digit in unit's place be 'x' and the digit in ten's place be 'y'.

According to the given condition,
the sum of a two digit number and the number
obtained by interchanging its digits is 99.
\( \therefore 10y + x + 10x + y = 99 \)
\( \therefore 11x + 11y = 99 \)
Dividing both sides by 11,
\( x + y = 9 \)
If \( y = 1 \), then \( x = 8 \)
If \( y = 2 \), then \( x = 7 \)
If \( y = 3 \), then \( x = 6 \) and so on.
\( \therefore \) The number can be 18, 27, 36, ... etc.
In simple words: By representing a two-digit number using its tens and units digits (y and x) and forming an equation based on the sum of the original number and the number with interchanged digits, we found that the sum of the digits must be 9, leading to multiple possible numbers like 18, 27, 36, etc.

🎯 Exam Tip: When dealing with two-digit number problems, remember to express the number as \( 10 \times \text{tens digit} + \text{units digit} \). This algebraic representation is crucial for setting up the correct equations.

Maharashtra Board Class 9 Chapter 5 Linear Equations in Two Variables Practice Set 5 Intext Questions And Activities

Question 1. On the glasses of following spectacles, write numbers such that (Textbook pg. no. 82)
i. Their sum is 42 and difference is 16.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक चश्मे को दर्शाता है जिसके प्रत्येक लेंस में एक संख्या (बाएं 29 और दाएं 13) लिखी है। ये संख्याएँ इस स्थिति को दर्शाती हैं कि उनका योग 42 है (29+13) और उनका अंतर 16 है (29-13)।
ii. Their sum is 37 and difference is 11.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक चश्मे को दर्शाता है जिसके प्रत्येक लेंस में कोई संख्या नहीं लिखी है। छात्रों को इस स्थिति के लिए उपयुक्त संख्याएँ (जिनका योग 37 और अंतर 11 हो) इन खाली लेंसों में लिखने को कहा गया है।
Answer: ii. \( x + y = 37 \) and \( x - y = 11 \)
\( \therefore x = 24, y = 13 \)
In simple words: For two numbers whose sum is 37 and difference is 11, solving the simultaneous equations \( x+y=37 \) and \( x-y=11 \) yields the numbers 24 and 13.

🎯 Exam Tip: Simple word problems involving sums and differences can be quickly solved by setting up two linear equations and using addition/subtraction to eliminate one variable.

iii. Their sum is 54 and difference is 20.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक चश्मे को दर्शाता है जिसके प्रत्येक लेंस में कोई संख्या नहीं लिखी है। छात्रों को इस स्थिति के लिए उपयुक्त संख्याएँ (जिनका योग 54 और अंतर 20 हो) इन खाली लेंसों में लिखने को कहा गया है।
iv. Their sum is ... and difference is ....
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक चश्मे को दर्शाता है जिसके प्रत्येक लेंस में कोई संख्या नहीं लिखी है। छात्रों को इस स्थिति के लिए उपयुक्त संख्याएँ और उनकी योग व अंतर की शर्तों को इन खाली लेंसों में लिखने को कहा गया है।

iii. \( x + y = 54 \) and \( x - y = 20 \)
\( \therefore x = 37, y = 17 \)

Question 2. There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs. (Textbook pg. no. 92)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयत को दर्शाता है जिसकी लंबाई 'x' इकाइयाँ और चौड़ाई 'y' इकाइयाँ हैं। आयत के चारों ओर तीरों के साथ चार अलग-अलग स्थितियाँ दी गई हैं: (1) लंबाई और चौड़ाई का योग 36 है (\( x+y=36 \)), (2) चौड़ाई लंबाई का \( \frac{5}{7} \) गुना है (\( y=\frac{5}{7}x \)), (3) लंबाई और चौड़ाई का अंतर 6 इकाइयाँ है (\( x-y=6 \)), और (4) यदि चौड़ाई को लंबाई के दोगुने से घटाया जाता है, तो उत्तर 27 है (\( 2x-y=27 \))। इन सभी शर्तों से संबंधित समीकरण बॉक्स में लिखे गए हैं।
Here, if we take a pair of any two equations, we get following 6 pairs.
1. equation (i) and (ii)
2. equation (i) and (iii)
3. equation (i) and (iv)
4. equation (ii) and (iii)
5. equation (ii) and (iv)
6. equation (iii) and (iv)
Solution of each pair given above is (21, 15).
Here, all four equations are of same rectangle. By solving any two equations simultaneously, we get length and breadth of the rectangle.
In simple words: We analyzed a diagram showing a rectangle's dimensions (x and y) and four conditions related to them, which led to four linear equations. Any pair of these equations can be solved simultaneously, forming 6 unique pairs. The solution for each pair is (21, 15), meaning the rectangle's length is 21 units and breadth is 15 units.

🎯 Exam Tip: Complex diagrams with multiple conditions can be simplified by breaking down each condition into a separate linear equation. Remember that all conditions for the same object should yield consistent solutions.

Question 3. Find the function.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रवाह चार्ट है जो एक भिन्न (fraction) को 'अंश x' और 'हर y' के रूप में दर्शाता है। इसमें दो स्थितियाँ दी गई हैं: पहली, यदि अंश को 3 से गुणा किया जाता है और हर से 3 घटाया जाता है, तो प्राप्त भिन्न \( \frac{18}{11} \) होता है (इससे समीकरण \( 11x-6y+18=0 \) मिलता है)। दूसरी, यदि अंश को 8 से बढ़ाया जाता है और हर को दोगुना किया जाता है, तो परिणामी भिन्न \( \frac{1}{2} \) होता है (इससे समीकरण \( x-y+8=0 \) मिलता है)।
Verify the answer obtained. (Textbook pg. no. 92)
For the fraction \( \frac{6}{14} \), if the numerator is multiplied by 3 and 3 is subtracted from the denominator, we get fraction \( \frac{18}{11} \).
Similarly, for the fraction \( \frac{6}{14} \), if the numerator is increased by 8 and the denominator is doubled, we get fraction \( \frac{1}{2} \).
In simple words: We are asked to find a fraction that satisfies two conditions related to changing its numerator and denominator. The fraction \( \frac{6}{14} \) is given as the answer, and it successfully verifies both conditions when tested.

🎯 Exam Tip: Problems involving fractions and their transformations often require careful algebraic manipulation. Expressing the fraction as \( \frac{x}{y} \) and then translating the conditions into equations is key.