Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 5 Linear Equations in Two Variables Set 5.1 Solutions

Question 1. By using variables x and y form any five linear equations in two variables.
Answer: The general form of a linear equation in two variables x and y is ax + by + c = 0, where a, b, c are real numbers and a \( \ne \) 0, b \( \ne \) 0.
Five linear equations in two variables are as follows:
(i) \( 3x + 4y - 12 = 0 \)
(ii) \( 3x - 4y + 12 = 0 \)
(iii) \( 5x + 5y - 6 = 0 \)
(iv) \( 7x + 12y - 11 = 0 \)
(v) \( x - y + 5 = 0 \)
In simple words: A linear equation in two variables forms a straight line when graphed. Any equation in the form ax + by + c = 0 (where a, b are not both zero) fits this description. You can pick any five such combinations for a, b, and c to create valid equations.

🎯 Exam Tip: Remember to specify that a and b should not both be zero, as this is a defining characteristic of a linear equation in two variables.

Question 2. Write five solutions of the equation x + y = 1.
Answer:
(i) x = 1, y = 6
(ii) x = -1, y = 8
(iii) x = 5, y = 2
(iv) x = 0, y = 7
(v) x = 10, y = -3
In simple words: A solution to an equation means a pair of values (x, y) that makes the equation true. For x + y = 1, you can choose any value for x, and then y will be 1 - x. For example, if x=1, then y=0 (1+0=1), if x=5, then y=-4 (5-4=1). The solutions provided in the answer are examples where x + y = 7 (e.g. 1+6=7, -1+8=7, 5+2=7, 0+7=7, 10-3=7), not x+y=1. Let me correct the answer to match the question's equation `x + y = 1`.
Answer:
(i) x = 1, y = 0
(ii) x = 2, y = -1
(iii) x = 0, y = 1
(iv) x = -1, y = 2
(v) x = 5, y = -4
In simple words: A solution to an equation means a pair of values (x, y) that makes the equation true. For x + y = 1, you can choose any value for x, and then y will be 1 - x. For example, if x = 1, then y = 0, because 1 + 0 = 1.

🎯 Exam Tip: When asked for multiple solutions, pick a variety of positive, negative, and zero values for one variable and calculate the corresponding value for the other variable.

Question 3. Solve the following sets of simultaneous equations.
(i) \( x + y = 4 \) ; \( 2x - 5y = 1 \)
(ii) \( 2x + y = 5 \) ; \( 3x - y = 5 \)
(iii) \( 3x - 5y = 16 \) ; \( x - 3y = 8 \)
(iv) \( 2y - x = 0 \) ; \( 10x + 15y = 105 \)
(v) \( 2x + 3y + 4 = 0 \) ; \( x - 5y = 11 \)
(vi) \( 2x - 7y = 7 \) ; \( 3x + y = 22 \)
Answer:
(i) Substitution Method:
\( x + y = 4 \)
\( \implies x = 4 - y \)...(i)
\( 2x - 5y = 1 \)......(ii)
Substituting \( x = 4 - y \) in equation (ii),
\( 2(4 - y) - 5y = 1 \)
\( \implies 8 - 2y - 5y = 1 \)
\( \implies 8 - 7y = 1 \)
\( \implies 8 - 1 = 7y \)
\( \implies 7 = 7y \)
\( \implies y = \frac{7}{7} \)
\( \implies y = 1 \)
Substituting \( y = 1 \) in equation (i),
\( x = 4 - 1 = 3 \)
\( \implies (3,1) \) is the solution of the given equations.
Alternate method:
Elimination Method:
\( x + y = 4 \)...(i)
\( 2x - 5y = 1 \)......(ii)
Multiplying equation (i) by 5,
\( 5x + 5y = 20 \)... (iii)
Adding equations (ii) and (iii),
\( 2x - 5y = 1 \)
\( + 5x + 5y = 20 \)
\( \implies 7x = 21 \)
\( \implies x = \frac{21}{7} \)
\( \implies x = 3 \)
Substituting \( x = 3 \) in equation (i),
\( 3 + y = 4 \)
\( \implies y = 4 - 3 = 1 \)
\( \implies (3,1) \) is the solution of the given equations.

(ii) \( 2x + y = 5 \)...(i)
\( 3x - y = 5 \)...(ii)
Adding equations (i) and (ii),
\( 2x + y = 5 \)
\( + 3x - y = 5 \)
\( \implies 5x = 10 \)
\( \implies x = \frac{10}{5} \)
\( \implies x = 2 \)
Substituting \( x = 2 \) in equation (i),
\( 2(2) + y = 5 \)
\( \implies 4 + y = 5 \)
\( \implies y = 5 - 4 = 1 \)
\( \implies (2, 1) \) is the solution of the given equations.

(iii) \( 3x - 5y = 16 \)...(i)
\( x - 3y = 8 \)
\( \implies x = 8 + 3y \).....(ii)
Substituting \( x = 8 + 3y \) in equation (i),
\( 3(8 + 3y) - 5y = 16 \)
\( \implies 24 + 9y - 5y = 16 \)
\( \implies 4y = 16 - 24 \)
\( \implies 4y = -8 \)
\( \implies y = \frac{-8}{4} \)
\( \implies y = -2 \)
Substituting \( y = -2 \) in equation (ii),
\( x = 8 + 3 (-2) \)
\( \implies x = 8 - 6 = 2 \)
\( \implies (2, -2) \) is the solution of the given equations.

(iv) \( 2y - x = 0 \)
\( \implies x = 2y \)...(i)
\( 10x + 15y = 105 \)...(ii)
Substituting \( x = 2y \) in equation (ii),
\( 10(2y) + 15y = 105 \)
\( \implies 20y + 15y = 105 \)
\( \implies 35y = 105 \)
\( \implies y = \frac{105}{35} \)
\( \implies y = 3 \)
Substituting \( y = 3 \) in equation (i),
\( x = 2y \)
\( \implies x = 2(3) = 6 \)
\( \implies (6, 3) \) is the solution of the given equations.

(v) \( 2x + 3y + 4 = 0 \)...(i)
\( x - 5y = 11 \)
\( \implies x = 11 + 5y \)...(ii)
Substituting \( x = 11 + 5y \) in equation (i),
\( 2(11 + 5y) + 3y + 4 = 0 \)
\( \implies 22 + 10y + 3y + 4 = 0 \)
\( \implies 13y + 26 = 0 \)
\( \implies 13y = -26 \)
\( \implies y = \frac{-26}{13} \)
\( \implies y = -2 \)
Substituting \( y = -2 \) in equation (ii),
\( x = 11 + 5y \)
\( \implies x = 11 + 5(-2) \)
\( \implies x = 11 - 10 = 1 \)
\( \implies (1, -2) \) is the solution of the given equations.

(vi) \( 2x - 7y = 7 \)...(i)
\( 3x + y = 22 \)
\( \implies y = 22 - 3x \)......(ii)
Substituting \( y = 22 - 3x \) in equation (i),
\( 2x - 7(22 - 3x) = 7 \)
\( \implies 2x - 154 + 21x = 7 \)
\( \implies 23x = 7 + 154 \)
\( \implies 23x = 161 \)
\( \implies x = \frac{161}{23} \)
\( \implies x = 7 \)
Substituting \( x = 7 \) in equation (ii),
\( y = 22 - 3x \)
\( \implies y = 22 - 3(7) \)
\( \implies y = 22 - 21 = 1 \)
\( \implies (7, 1) \) is the solution of the given equations.
In simple words: Simultaneous equations mean finding a single pair of (x, y) values that satisfies both equations at the same time. This can be done by substitution (expressing one variable in terms of the other) or elimination (multiplying equations to make coefficients of one variable equal and then adding or subtracting).

🎯 Exam Tip: Always verify your solution by substituting the calculated x and y values back into both original equations to ensure they hold true. This catches errors.

Question 1. Solve the following equations.
Answer:
(i) \( m + 3 = 5 \)
\( \implies m = 5 - 3 \)
\( \implies m = 2 \)

(ii) \( 3y + 8 = 22 \)
\( \implies 3y = 22 - 8 \)
\( \implies 3y = 14 \)
\( \implies y = \frac{14}{3} \)

(iii) \( \frac{x}{3} = 2 \)
\( \implies x = 2 \times 3 \)
\( \implies x = 6 \)

(iv) \( 2p = p + \frac{4}{9} \)
\( \implies 2p - p = \frac{4}{9} \)
\( \implies p = \frac{4}{9} \)
In simple words: These are basic linear equations with one variable. To solve them, isolate the variable by performing inverse operations (addition/subtraction, multiplication/division) on both sides of the equation.

🎯 Exam Tip: When transposing terms across the equals sign, remember to change their operation (e.g., addition becomes subtraction, multiplication becomes division).

Question 2. Which number should be added to 5 to obtain 14?
Answer:
Let the number be \( x \).
\( x + 5 = 14 \)
\( \implies x = 14 - 5 \)
\( \implies x = 9 \)
\( \implies 9 + 5 = 14 \)
In simple words: To find the unknown number, set up an equation where 5 plus the unknown number equals 14, then solve for the unknown by subtracting 5 from 14.

🎯 Exam Tip: Clearly define your unknown variable at the beginning of word problems to make your solution easy to follow.

Question 3. Which number should be subtracted from 8 to obtain 2?
Answer:
Let the number be \( y \).
\( 8 - y = 2 \)
\( \implies y = 8 - 2 \)
\( \implies y = 6 \)
\( \implies 8 - 6 = 2 \)
In simple words: Represent the unknown number as a variable. Set up an equation where 8 minus that variable equals 2, then solve for the variable.

🎯 Exam Tip: Be careful with the order of subtraction; "subtracted from 8" means 8 - y, not y - 8.

Question 4. \( x + y = 5 \) and \( 2x + 2y = 10 \) are two equations in two variables. Find five different solutions of \( x + y = 5 \), verify whether same solutions satisfy the equation \( 2x + 2y = 10 \) also. Observe both equations. Find the condition where two equations in two variables have all solutions in common.
Answer:
Five solutions of \( x + y = 5 \) are given below:
\( (1,4), (2, 3), (3, 2), (4,1), (0, 5) \)
The above solutions also satisfy the equation \( 2x + 2y = 10 \).
\( \implies x + y = 5 \) ...[Dividing both sides by 2]
\( \implies \) If the two equations are the same, then the two equations in two variables have all solutions common.
In simple words: If two linear equations are essentially the same (one is a multiple of the other), they represent the same line and thus share all their solutions, meaning they have infinitely many common solutions.

🎯 Exam Tip: Recognize that equations which are scalar multiples of each other (e.g., \( 2x+2y=10 \) is \( 2 \times (x+y=5) \)) are dependent and represent the same line, hence having infinite solutions.

Question 5. \( 3x - 4y - 15 = 0 \) and \( y + x + 2 = 0 \). Can these equations be solved by eliminating x ? Is the solution same?
Answer:
\( 3x - 4y - 15 = 0 \)
\( \implies 3x - 4y = 15 \)...(i)
\( y + x + 2 = 0 \)
\( \implies x + y = -2 \)......(ii)
Multiplying equation (ii) by 3,
\( 3x + 3y = -6 \)...(iii)
Subtracting equation (iii) from (i),
\( 3x - 4y = 15 \)
\( -(3x + 3y = -6) \)
\( \implies -7y = 21 \)
\( \implies y = \frac{-21}{7} \)
\( \implies y = -3 \)
Substituting \( y = -3 \) in equation (ii),
\( x + (-3) = -2 \)
\( \implies x - 3 = -2 \)
\( \implies x = -2 + 3 \)
\( \implies x = 1 \)
\( \implies (x, y) = (1, -3) \)
Yes, the given equations can be solved by eliminating x. Also, the solution will remain the same.
In simple words: You can solve these equations by elimination. Multiply the second equation by 3 to make the 'x' coefficients equal, then subtract the equations to eliminate 'x' and solve for 'y'. Substitute 'y' back into one of the original equations to find 'x'. The solution obtained will be unique and valid.

🎯 Exam Tip: Always write the equations in standard form (ax + by = c) before attempting elimination or substitution. Be careful with signs when subtracting equations.