Q3(iv):

Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (iv) $(\sqrt{2}, 4\sqrt{2})$

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.

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.

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 $x - 2y = 4$.