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Discover comprehensive, step-by-step solutions for Class 9 Maths Chapter 4, Exercise 4.4, which continues the exploration of Linear Equations in Two Variables with a focus on practical applications and problem-solving. This exercise challenges students to apply their knowledge to real-life situations, interpreting given equations and drawing graphs accordingly. Through solving these problems, learners enhance their ability to analyze relationships between variables, construct tables of values, and understand how linear equations model real-world scenarios. Exercise 4.4 builds on graphical representation skills and encourages logical thinking—essential for mastering advanced algebra and applied mathematics.
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Exercise 4.4
(i) In one variable
(ii) In two variables
For that, we can represent a point marked on the number line. And for \( y =3 \), mark y at \( {y}= \)
(ii) In two variables, it is represented as a line that is parallel to \(X\) -axis \( 0 x+y=3 \), which shows that the \( y \) component of the graph will always remain constant and will be equal to 3.
So it will be a straight line parallel to \(x\) -axis and \(y\) coordinate equal to 0
2. Give the geometric representations of \( 2 {x}+9=0 \) as an equation
\(2 x+9=0\)
or, \( 2 {x}=-9 \)
\(\text {or, } x=-\frac{ 9 }{ 2 }\)
In one variable, it can be represented on a straight line with one dimension or coordinate only, so it will be represented as \( x=-\frac{ 9 }{ 2 } \)
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Now if we put \( {y}=02 {x}+9=0 {x}=\frac{ -9 }{ 2 } \) If we put \( {y}=12 {x}+0(1)+9=02 {x} \) \( +9=0 x=-\frac{ 9 }{ 2 } \)
Thus it will be a line having a constant value of \( x \) that is \( -\frac{ 9 }{ 2 } \) and the value can be anything, thus a straight line parallel to the \( y \)-axis.
