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Q3(iv):
Check which of the following are solutions of the equation $x - 2y = 4$ and which are not:
(iv) $(\sqrt{2}, 4\sqrt{2})$
Solution :
Initial Setup & Given Variables
We are tasked with verifying whether the coordinate pair $(\sqrt{2}, 4\sqrt{2})$ satisfies the given linear equation in two variables. The foundational components of our analysis are:
- The Linear Equation: $x - 2y = 4$
- The Ordered Pair: $(x, y) = (\sqrt{2}, 4\sqrt{2})$
[Per the fundamental theorem of algebra and coordinate geometry, an ordered pair $(x, y)$ is a valid solution to an equation if and only if substituting the values of $x$ and $y$ into the equation results in a true mathematical statement where the Left Hand Side (LHS) equals the Right Hand Side (RHS)].
Step 1: Substitution of Coordinates
We isolate the Left Hand Side (LHS) of the equation and substitute the specific values from the ordered pair.
Given LHS: $x - 2y$
Substitute $x = \sqrt{2}$ and $y = 4\sqrt{2}$:
$\text{LHS} = (\sqrt{2}) - 2(4\sqrt{2})$
Step 2: Algebraic Evaluation
Next, we simplify the expression using the properties of real numbers and radical arithmetic.
Multiply the constant $2$ by the coefficient of the radical term $4\sqrt{2}$ [By the associative property of multiplication]:
$\text{LHS} = \sqrt{2} - 8\sqrt{2}$
Factor out the common radical term $\sqrt{2}$ [By the distributive property of multiplication over addition/subtraction]:
$\text{LHS} = (1 - 8)\sqrt{2}$
$\text{LHS} = -7\sqrt{2}$
Step 3: Comparison with Right Hand Side (RHS)
We now compare the evaluated LHS with the constant RHS of the original equation.
| Left Hand Side (LHS) | Right Hand Side (RHS) | Logical Relation |
|---|---|---|
| $-7\sqrt{2}$ | $4$ | $\text{LHS} \neq \text{RHS}$ |
Since $-7\sqrt{2}$ is an irrational number approximately equal to $-9.899$, and the RHS is the rational integer $4$, the two sides are strictly unequal.
Geometric Verification
Geometrically, the equation $x - 2y = 4$ represents a straight line on the Cartesian plane. If the point $(\sqrt{2}, 4\sqrt{2})$ were a solution, it would lie exactly on this line. The precise SVG rendering below demonstrates the spatial divergence between the line and the coordinate point.
Figure 1: The point $(\sqrt{2}, 4\sqrt{2})$ clearly does not intersect the line $x - 2y = 4$.
Final Solution: Since substituting $x = \sqrt{2}$ and $y = 4\sqrt{2}$ yields $-7\sqrt{2} \neq 4$, the ordered pair $(\sqrt{2}, 4\sqrt{2})$ is NOT a solution to the equation $x - 2y = 4$.
More Questions from Class 9 Mathematics Linear Equations in Two Variables EXERCISE 4.2
- Q1: Which one of the following options is true, and why? $y = 3x + 5$ has
- Q2(i): Write four solutions for each of the following equations: (i) $2x + y = 7$
- Q2(ii): Write four solutions for each of the following equations: (ii) $\pi x + y = 9$
- Q2(iii): Write four solutions for each of the following equations: (iii) $x = 4y$
- Q3(i): Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (i) $(0, 2)$
- Q3(ii): Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (ii) $(2, 0)$
- Q3(iii): Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (iii) $(4, 0)$
- Q3(v): Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (v) $(1, 1)$
- Q4: Find the value of $k$, if $x = 2$, $y = 1$ is a solution of the equation $2x + 3y = k$.
CBSE Solutions for Class 9 Mathematics Linear Equations in Two Variables
Chapters in CBSE - Class 9 Mathematics
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