GSEB Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 4.1

Question 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Answer: Let the cost of a notebook be \( x \) and the cost of a pen be \( y \). According to the given problem, the cost of the notebook is double the pen's cost.
\( x = 2y \)
\( \implies x - 2y = 0 \)
This is the required equation that expresses the statement in two variables, \( x \) and \( y \).
In simple words: If a notebook costs \( x \) and a pen costs \( y \), and the notebook is twice as expensive, the equation is \( x = 2y \), which can also be written as \( x - 2y = 0 \).

Exam Tip: Clearly define the variables you use (e.g., let x be the cost of the notebook) before forming the equation. Always simplify the equation to the standard form \( ax + by + c = 0 \) when possible.

Question 2. Express the following linear equations in the form \( ax + by + c = 0 \) and indicate the values of a, b and c in each case.
(i) \( 2x + 3y = 9.35 \)
(ii) \( x - y - 10 = 0 \)
(iii) \( -2x + 3y = 6 \)
(iv) \( x = 3y \)
(v) \( 2x = -5y \)
(vi) \( 3x + 2 = 0 \)
(vii) \( y - 2 = 0 \)
(viii) \( 5 = 2x \)
Answer:
(i) Given equation is \( 2x + 3y = 9.35 \).
To express it in the form \( ax + by + c = 0 \), we move the constant term to the left side:
\( 2x + 3y - 9.35 = 0 \)
Comparing this with \( ax + by + c = 0 \), we get:
\( a = 2, b = 3, c = -9.35 \)

(ii) Given equation is \( x - y - 10 = 0 \).
This equation is already in the standard form \( ax + by + c = 0 \).
We can explicitly write it as \( 1x + (-1)y + (-10) = 0 \).
Comparing this with \( ax + by + c = 0 \), we get:
\( a = 1, b = -1, c = -10 \)

(iii) Given equation is \( -2x + 3y = 6 \).
To express it in the form \( ax + by + c = 0 \), we transfer the constant term to the left side:
\( -2x + 3y - 6 = 0 \)
We can also write it as \( (-2)x + 3y + (-6) = 0 \).
Comparing this with \( ax + by + c = 0 \), we find the values:
\( a = -2, b = 3, c = -6 \)

(iv) Given equation is \( x = 3y \).
To express it in the form \( ax + by + c = 0 \), we move all terms to one side:
\( x - 3y = 0 \)
We can write this as \( 1x + (-3)y + 0 = 0 \).
Comparing this with \( ax + by + c = 0 \), we identify the coefficients:
\( a = 1, b = -3, c = 0 \)

(v) Given equation is \( 2x = -5y \).
To express it in the form \( ax + by + c = 0 \), we relocate all terms to one side:
\( 2x + 5y = 0 \)
We can write this as \( 2x + 5y + 0 = 0 \).
Comparing this with \( ax + by + c = 0 \), we determine the values:
\( a = 2, b = 5, c = 0 \)

(vi) Given equation is \( 3x + 2 = 0 \).
To express it in the form \( ax + by + c = 0 \), we can include the \( y \) term with a zero coefficient:
\( 3x + 0y + 2 = 0 \)
Comparing this with \( ax + by + c = 0 \), we obtain:
\( a = 3, b = 0, c = 2 \)

(vii) Given equation is \( y - 2 = 0 \).
To express it in the form \( ax + by + c = 0 \), we can incorporate the \( x \) term with a zero coefficient:
\( 0x + 1y + (-2) = 0 \)
Comparing this with \( ax + by + c = 0 \), we get:
\( a = 0, b = 1, c = -2 \)

(viii) Given equation is \( 5 = 2x \).
To express it in the form \( ax + by + c = 0 \), we rearrange the terms so that all terms are on one side, with the \( x \) term first:
\( -2x + 5 = 0 \)
We can then write this as \( (-2)x + 0y + 5 = 0 \).
Comparing this with \( ax + by + c = 0 \), we find the coefficients:
\( a = -2, b = 0, c = 5 \)
In simple words: For each equation, move all terms to one side to make it equal zero. Then, match it to the form \( ax + by + c = 0 \) to see what numbers \( a \), \( b \), and \( c \) stand for. If a variable is missing, its coefficient (a or b) is zero.

Exam Tip: Remember to correctly identify the signs (positive or negative) of a, b, and c when comparing with the standard form \( ax + by + c = 0 \). Always include a zero coefficient for any missing variable.