Initial Setup & Theoretical Foundation

We are given the linear equation:

$2x = -5y$

The objective is to express this equation in the general (standard) form of a linear equation in two variables, which is defined as:

$ax + by + c = 0$

where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero simultaneously (i.e., $a^2 + b^2 \neq 0$).

Step 1: Algebraic Transposition

To convert the given equation into the standard form, all non-zero terms must be collected on the left-hand side (LHS) of the equation, leaving zero on the right-hand side (RHS).

We apply the Addition Property of Equality [which states that adding the same quantity to both sides of an equation preserves the equality]. We add $5y$ to both sides of the equation:

$2x + 5y = -5y + 5y$

$2x + 5y = 0$

Step 2: Structuring the Standard Form

The standard form requires three distinct terms: an $x$-term, a $y$-term, and a constant term $c$. In our current equation, $2x + 5y = 0$, the constant term is absent. Mathematically, an absent term implies a coefficient or value of zero. We explicitly introduce zero as the constant term to perfectly match the $ax + by + c = 0$ template:

$2x + 5y + 0 = 0$

Step 3: Coefficient Extraction

By directly comparing our structured equation with the general form $ax + by + c = 0$, we can extract the corresponding real number coefficients:

• The coefficient of $x$ corresponds to $a$. Thus, $a = 2$.
• The coefficient of $y$ corresponds to $b$. Thus, $b = 5$.
• The constant term corresponds to $c$. Thus, $c = 0$.

Graphical Representation (Geometric Interpretation)

A linear equation in two variables geometrically represents a straight line on a Cartesian plane. Because the constant term $c = 0$, the line must pass through the origin $(0,0)$. We can verify this by finding two points that satisfy $2x + 5y = 0$:

• If $x = 0$, then $5y = 0 \implies y = 0$. Point: $(0, 0)$
• If $x = 5$, then $10 + 5y = 0 \implies 5y = -10 \implies y = -2$. Point: $(5, -2)$
• If $x = -5$, then $-10 + 5y = 0 \implies 5y = 10 \implies y = 2$. Point: $(-5, 2)$

Final Solution: The linear equation expressed in standard form is $2x + 5y + 0 = 0$. The corresponding coefficient values are $a = 2$, $b = 5$, and $c = 0$.

Initial Setup & Given Equation

We are given the linear equation:

$3x + 2 = 0$

The objective is to express this equation in the standard two-variable form, $ax + by + c = 0$, and to identify the corresponding real coefficients $a$, $b$, and $c$.

Step 1: Algebraic Transformation to Standard Form

The standard form of a linear equation in two variables is defined as:

$ax + by + c = 0$

where $x$ and $y$ are variables, and $a$, $b$, and $c$ are real numbers such that $a$ and $b$ are not simultaneously zero [Per the fundamental definition of a linear equation in two variables].

Observing the given equation $3x + 2 = 0$, we note the absence of the $y$-variable. To represent this equation in a two-variable format without altering its mathematical equivalence, we introduce the variable $y$ with a coefficient of $0$. [Since $0 \cdot y = 0$ for any real number $y$, the additive identity property ensures the equation's value remains unchanged].

Rewriting the equation yields:

$3x + 0 \cdot y + 2 = 0$

Step 2: Extraction of Coefficients

By aligning our transformed equation directly with the standard form, we can map the corresponding coefficients:

• Standard Form: $ax + by + c = 0$
• Transformed Equation: $3x + 0y + 2 = 0$

Comparing the terms positionally:

• The coefficient of $x$ is $a = 3$.
• The coefficient of $y$ is $b = 0$.
• The constant term is $c = 2$.

Geometric Interpretation (Graphical Representation)

Geometrically, solving the equation $3x + 2 = 0$ for $x$ yields $x = -\frac{2}{3}$. In a two-dimensional Cartesian coordinate system, an equation of the form $x = k$ (where $k$ is a constant) represents a vertical line parallel to the $y$-axis. The line intersects the $x$-axis at the coordinate $(-\frac{2}{3}, 0)$. Because $y$ can take any real value while $x$ remains constant, the coefficient of $y$ is effectively zero.

Final Conclusion

Final Solution: The equation expressed in the standard form $ax + by + c = 0$ is $3x + 0y + 2 = 0$. The corresponding coefficients are $a = 3$, $b = 0$, and $c = 2$.

Initial Setup & Theoretical Foundation

We are given the linear equation in two variables:

$x = 4y$

[Per the fundamental theorem of linear algebra], a linear equation in two variables of the form $ax + by + c = 0$ has infinitely many solutions. This is because for every arbitrary real value assigned to the independent variable, there exists a unique corresponding real value for the dependent variable. To find four distinct solutions, we will systematically assign four different real values to $y$ and solve for the corresponding values of $x$.

Step 1: Deriving the First Solution

Let us assign the value $y = 0$. Substituting this into the given equation:

$x = 4(0)$

$x = 0$

Thus, the first ordered pair $(x, y)$ that satisfies the equation is $(0, 0)$.

Step 2: Deriving the Second Solution

Let us assign the value $y = 1$. Substituting this into the given equation:

$x = 4(1)$

$x = 4$

Thus, the second ordered pair $(x, y)$ that satisfies the equation is $(4, 1)$.

Step 3: Deriving the Third Solution

Let us assign the value $y = 2$. Substituting this into the given equation:

$x = 4(2)$

$x = 8$

Thus, the third ordered pair $(x, y)$ that satisfies the equation is $(8, 2)$.

Step 4: Deriving the Fourth Solution

Let us assign a negative integer to demonstrate solutions in the third quadrant. Let $y = -1$. Substituting this into the given equation:

$x = 4(-1)$

$x = -4$

Thus, the fourth ordered pair $(x, y)$ that satisfies the equation is $(-4, -1)$.

Tabular Representation of Solutions

We can organize the derived solutions into a coordinate table to clearly map the relationship between $x$ and $y$.

Graphical Verification

Plotting these points on a Cartesian plane confirms that they are collinear, forming the straight line defined by $x = 4y$.

Final Solution: Four valid solutions for the equation $x = 4y$ are $(0, 0)$, $(4, 1)$, $(8, 2)$, and $(-4, -1)$.

Initial Setup & Algebraic Formulation

We are tasked with finding four distinct solutions for the following linear equation in two variables:

$\pi x + y = 9$

A linear equation in two variables has infinitely many solutions. A solution is defined as an ordered pair $(x, y)$ that satisfies the equation. To systematically generate these solutions, we will express one variable in terms of the other.

Step 1: Isolating the Dependent Variable

To make the calculation of ordered pairs straightforward, we isolate the dependent variable ($y$) in terms of the independent variable ($x$). [Per the Subtraction Property of Equality, subtracting $\pi x$ from both sides of the equation yields:]

$y = 9 - \pi x$

We can now substitute arbitrary real numbers for $x$ to compute the corresponding exact values of $y$.

Step 2: Deriving the First Solution

Let us assign $x = 0$. Substituting this value into our isolated equation:

$y = 9 - \pi(0)$

$y = 9 - 0$

$y = 9$

Thus, the first solution is the ordered pair $(0, 9)$. [This point also represents the y-intercept of the linear graph].

Step 3: Deriving the Second Solution

Let us assign $x = 1$. Substituting this value into the equation:

$y = 9 - \pi(1)$

$y = 9 - \pi$

Thus, the second solution is the ordered pair $(1, 9 - \pi)$.

Step 4: Deriving the Third Solution

Let us assign $x = 2$. Substituting this value into the equation:

$y = 9 - \pi(2)$

$y = 9 - 2\pi$

Thus, the third solution is the ordered pair $(2, 9 - 2\pi)$.

Step 5: Deriving the Fourth Solution

Let us assign $x = -1$. Substituting this value into the equation:

$y = 9 - \pi(-1)$

$y = 9 + \pi$

Thus, the fourth solution is the ordered pair $(-1, 9 + \pi)$.

Step 6: Tabular Summary of Solutions

We can organize the derived solutions into a Cartesian coordinate table. This demonstrates the linear relationship and prepares the data for graphical representation.

Step 7: Graphical Representation of the Linear Equation

Below is the geometric interpretation of the equation $\pi x + y = 9$ plotted on a Cartesian plane. The line represents the locus of all infinite solutions, with our four specific calculated points highlighted. (Note: For spatial plotting purposes, $\pi$ is approximated as $3.14$).

Final Solution: Four valid solutions for the equation $\pi x + y = 9$ are $(0, 9)$, $(1, 9 - \pi)$, $(2, 9 - 2\pi)$, and $(-1, 9 + \pi)$.

Initial Setup & Theoretical Foundation

We are given the linear equation:

$y - 2 = 0$

The objective is to express this equation in the standard general form of a linear equation in two variables, which is defined as:

$ax + by + c = 0$

[Per the fundamental theorem of linear algebra in two dimensions, $x$ and $y$ are the variables, while $a$, $b$, and $c$ are real numbers known as coefficients and constants, respectively. For the equation to represent a line in a two-dimensional Cartesian plane, $a$ and $b$ cannot both be simultaneously zero ($a^2 + b^2 \neq 0$)].

Step 1: Algebraic Manipulation to Standard Form

Observing the given equation $y - 2 = 0$, we note that the variable $x$ is absent. To map this to the two-variable standard form $ax + by + c = 0$, we must introduce the variable $x$ without altering the mathematical truth or value of the original equation.

[By the Zero Property of Multiplication, any real number multiplied by zero equals zero. Therefore, we can introduce the term $0 \cdot x$].

Rewriting the equation with the $x$ term:

$0 \cdot x + y - 2 = 0$

To make the coefficients explicitly clear, we can write the implied coefficient of $y$ (which is $1$) and express the subtraction of $2$ as the addition of a negative constant:

$0x + 1y + (-2) = 0$

Step 2: Extraction of Coefficients

Now, we perform a direct term-by-term comparison between our expanded equation and the standard form:

• Standard Form: $ax + by + c = 0$
• Expanded Equation: $0x + 1y + (-2) = 0$

By equating the corresponding coefficients, we derive:

• The coefficient of $x$ is $a \implies a = 0$
• The coefficient of $y$ is $b \implies b = 1$
• The constant term is $c \implies c = -2$

Geometric Interpretation (Visual Aid)

Geometrically, a linear equation where $a = 0$ and $b \neq 0$ represents a horizontal line parallel to the x-axis. The equation $0x + 1y - 2 = 0$ simplifies to $y = 2$. This means that for any real value of $x$, the value of $y$ remains strictly $2$.

Final Solution: The equation expressed in the standard form $ax + by + c = 0$ is $0x + 1y - 2 = 0$. The corresponding values of the coefficients are $a = 0$, $b = 1$, and $c = -2$.

Given Equation & Initial Setup

We are given the following linear equation in one variable:

$5 = 2x$

Our objective is to express this equation in the general standard form of a linear equation in two variables, which is given by:

$ax + by + c = 0$

where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero ($a^2 + b^2 \neq 0$).

Step 1: Algebraic Manipulation to Standard Form

To convert the given equation into the standard form, we must transpose all terms to one side of the equality so that the other side becomes zero. [Per the properties of equality, subtracting the same value from both sides preserves the truth value of the equation].

Subtracting $5$ from both sides of the equation:

$5 - 5 = 2x - 5$

$0 = 2x - 5$

By the symmetric property of equality ($A = B \implies B = A$), we can rewrite this as:

$2x - 5 = 0$

Note: Alternatively, subtracting $2x$ from both sides yields $-2x + 5 = 0$. Both forms are mathematically valid, but maintaining a positive leading coefficient is standard convention.

Step 2: Introducing the Second Variable

The standard form requires two variables, $x$ and $y$. Since the variable $y$ is absent from our current equation, it implies that the coefficient of $y$ is exactly zero. [Any real number multiplied by zero is zero, hence $0 \cdot y = 0$].

We explicitly introduce the $y$ term to match the $ax + by + c = 0$ structure:

$2x + 0y - 5 = 0$

Step 3: Identification of Coefficients

Now, we align our derived equation with the general standard form to extract the values of the constants $a$, $b$, and $c$:

• Standard Form: $ax + by + c = 0$
• Derived Equation: $2x + 0y + (-5) = 0$

By direct comparison of the corresponding terms:

• The coefficient of $x$ is $a \implies a = 2$
• The coefficient of $y$ is $b \implies b = 0$
• The constant term is $c \implies c = -5$

Geometric Representation (Graphical Analysis)

A linear equation in two variables where one coefficient is zero represents a line parallel to one of the coordinate axes. The equation $2x - 5 = 0$ simplifies to $x = 2.5$. This represents a vertical line parallel to the Y-axis, intersecting the X-axis at exactly $(2.5, 0)$.

Final Solution: The equation expressed in the standard form $ax + by + c = 0$ is $2x + 0y - 5 = 0$. The corresponding values of the coefficients are $a = 2$, $b = 0$, and $c = -5$.

(Note: The alternative valid form $-2x + 0y + 5 = 0$ yields $a = -2$, $b = 0$, $c = 5$.)

Initial Setup & Theoretical Foundation

The general standard form of a linear equation in two variables is defined algebraically as:

$ax + by + c = 0$

where $a$, $b$, and $c$ are real numbers ($a, b, c \in \mathbb{R}$), and the coefficients $a$ and $b$ are not simultaneously zero (often denoted as $a^2 + b^2 \neq 0$).

We are given the following linear equation:

$2x + 3y = 9.3\overline{5}$

Step 1: Transposition to Standard Form

To map the given equation to the standard form, we must collect all terms on the left-hand side of the equality, leaving exactly zero on the right-hand side.

[Per the Subtraction Property of Equality], we subtract the constant term $9.3\overline{5}$ from both sides of the equation. Note that $9.3\overline{5}$ is a rational real number (a non-terminating, repeating decimal), and it is treated algebraically as a single constant value.

$2x + 3y - 9.3\overline{5} = 9.3\overline{5} - 9.3\overline{5}$

$2x + 3y - 9.3\overline{5} = 0$

Step 2: Coefficient Extraction and Comparative Analysis

To strictly match the $ax + by + c = 0$ structure, we can conceptually express the subtraction of the constant as the addition of a negative constant:

$2x + 3y + (-9.3\overline{5}) = 0$

By applying the principle of polynomial identity, we equate the corresponding coefficients from our derived equation directly to the standard form variables:

• The coefficient of the $x$-term corresponds to $a \implies a = 2$
• The coefficient of the $y$-term corresponds to $b \implies b = 3$
• The constant term corresponds to $c \implies c = -9.3\overline{5}$

Final Solution: The linear equation expressed in standard form is $2x + 3y - 9.3\overline{5} = 0$. The corresponding coefficient values are $a = 2$, $b = 3$, and $c = -9.3\overline{5}$.

Initial Setup & Theoretical Foundation

The standard form of a linear equation in two variables is defined universally as:

$ax + by + c = 0$

Where:

• $x$ and $y$ are the variables.
• $a$, $b$, and $c$ are real numbers ($\mathbb{R}$).
• $a$ and $b$ are not both zero (i.e., $a^2 + b^2 \neq 0$).

We are given the linear equation:

$x - \frac{y}{5} - 10 = 0$

Step 1: Structural Alignment of the Equation

To express the given equation strictly in the form $ax + by + c = 0$, we must isolate the numerical coefficients for each term. We rewrite the equation by explicitly stating the multiplier for each variable and converting subtractions into the addition of negative quantities [Per the definition of algebraic subtraction: $A - B = A + (-B)$].

• The term $x$ can be written with an explicit coefficient of $1$: $1 \cdot x$
• The term $-\frac{y}{5}$ can be factored into a fractional coefficient multiplied by $y$: $\left(-\frac{1}{5}\right)y$
• The constant term $-10$ can be written as an added negative integer: $+ (-10)$

Substituting these explicit forms back into the equation yields:

$1x + \left(-\frac{1}{5}\right)y + (-10) = 0$

Step 2: Extraction of Coefficients

By directly comparing our structurally aligned equation to the standard form $ax + by + c = 0$, we establish a one-to-one correspondence between the coefficients:

Step 3: Alternative Integer Representation (Rigorous Extension)

While the fractional coefficients extracted above are perfectly correct, linear equations are frequently scaled to clear denominators for computational simplicity. By applying the Multiplication Property of Equality, we can multiply the entire equation by $5$ (the Least Common Multiple of the denominators):

$5 \cdot \left(x - \frac{y}{5} - 10\right) = 5 \cdot (0)$

$5x - y - 50 = 0$

Rewriting this in explicit standard form gives $5x + (-1)y + (-50) = 0$. In this equivalent integer-based representation, the coefficients would be $a = 5$, $b = -1$, and $c = -50$. Both sets of coefficients define the exact same geometric line.

Graphical Analysis of the Linear Equation

To visualize the geometric truth of this equation, we plot the line $x - \frac{y}{5} - 10 = 0$. Solving for $y$ yields the slope-intercept form $y = 5x - 50$. The line intersects the Cartesian axes at $(10, 0)$ and $(0, -50)$.

Final Solution: The equation expressed in the standard form $ax + by + c = 0$ is $1x + \left(-\frac{1}{5}\right)y + (-10) = 0$. The corresponding values for the coefficients are $a = 1$, $b = -\frac{1}{5}$, and $c = -10$.

Given Variables & Initial Setup

We are given the following linear equation in two variables:

$-2x + 3y = 6$

The objective is to express this equation in the general (standard) form of a linear equation in two variables, which is defined as:

$ax + by + c = 0$

where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero (i.e., $a^2 + b^2 \neq 0$).

Step 1: Algebraic Manipulation to Standard Form

To convert the given equation into the standard form, we must ensure that all terms (variables and constants) are collected on the left-hand side (LHS) of the equation, leaving exactly zero on the right-hand side (RHS).

• Original Equation: $-2x + 3y = 6$
• Transformation: Apply the Subtraction Property of Equality [Subtracting $6$ from both sides of the equation to nullify the RHS].
• $-2x + 3y - 6 = 6 - 6$
• $-2x + 3y - 6 = 0$

Step 2: Identification of Coefficients

Now, we align our transformed equation with the standard theoretical model to extract the corresponding coefficients.

Standard Form: $ax + by + c = 0$

Derived Equation: $(-2)x + (3)y + (-6) = 0$

By direct comparison of the corresponding terms, we identify the values:

• The coefficient of $x$, denoted as $a$, is $-2$.
• The coefficient of $y$, denoted as $b$, is $3$.
• The constant term, denoted as $c$, is $-6$.

Note on Mathematical Equivalence: Multiplying the entire equation by $-1$ yields $2x - 3y + 6 = 0$. In this equivalent standard form, the coefficients would be $a = 2$, $b = -3$, and $c = 6$. Both representations are mathematically rigorous and correct, though the direct transposition (yielding $a = -2$) is the primary expected procedure.

Graphical Representation of the Linear Equation

To provide a complete analytical perspective, the equation $-2x + 3y = 6$ represents a straight line on a Cartesian plane. The intercepts are calculated as follows:

• x-intercept: Set $y = 0 \implies -2x = 6 \implies x = -3$. Coordinate: $(-3, 0)$
• y-intercept: Set $x = 0 \implies 3y = 6 \implies y = 2$. Coordinate: $(0, 2)$

Final Solution: The linear equation expressed in the standard form $ax + by + c = 0$ is $-2x + 3y - 6 = 0$. The corresponding values of the coefficients are $a = -2$, $b = 3$, and $c = -6$.

Initial Setup & Objective

We are given the following linear equation in two variables:

$x = 3y$

The objective is to express this equation in the general standard form of a linear equation in two variables, which is given by:

$ax + by + c = 0$

where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero ($a^2 + b^2 \neq 0$). After transforming the equation, we must identify the specific numerical values for the coefficients $a$, $b$, and the constant term $c$.

Step 1: Algebraic Manipulation to Standard Form

To convert the given equation into the standard form, all non-zero terms must be transposed to one side of the equality, leaving a zero on the opposite side.

Given:

$x = 3y$

We subtract $3y$ from both sides of the equation [Per the Subtraction Property of Equality, which states that if $A = B$, then $A - C = B - C$]:

$x - 3y = 3y - 3y$

$x - 3y = 0$

Step 2: Explicit Representation of Coefficients

The standard form requires explicit representation of the $x$-term, the $y$-term, and the constant term $c$. We can rewrite the expression $x - 3y = 0$ by explicitly stating the hidden multiplicative identity (1) and the additive identity (0) [By the definition of the standard form of a linear equation]:

$1 \cdot x + (-3) \cdot y + 0 = 0$

Step 3: Identification of Parameters

By directly comparing our expanded equation to the standard form $ax + by + c = 0$, we can map the corresponding coefficients:

• The coefficient of $x$ is $a$. Therefore, $a = 1$.
• The coefficient of $y$ is $b$. Therefore, $b = -3$.
• The constant term is $c$. Since there is no constant term present in the original equation, $c = 0$.

Note: An alternative, equally valid standard form can be achieved by subtracting $x$ from both sides (yielding $-x + 3y + 0 = 0$), which would result in $a = -1$, $b = 3$, and $c = 0$. However, it is conventional mathematical practice to maintain a positive leading coefficient for $x$.

Graphical Representation & Analysis

Because the constant term $c = 0$, the linear equation $x = 3y$ (or $y = \frac{1}{3}x$) represents a straight line that passes directly through the origin $(0,0)$. For every 3 units moved along the x-axis, the line moves 1 unit along the y-axis.

Final Solution: The linear equation expressed in the standard form $ax + by + c = 0$ is $1x - 3y + 0 = 0$. The corresponding values for the coefficients are $a = 1$, $b = -3$, and $c = 0$.

Step 1: Initial Setup & Given Variables

We are given the linear equation in two variables:

$x - 2y = 4$

We need to determine whether the coordinate point $(0, 2)$ is a valid solution to this equation. In the Cartesian coordinate system, an ordered pair $(x, y)$ represents a solution to an equation if and only if substituting the $x$ and $y$ values into the equation results in a true mathematical statement [Per the Fundamental Principle of Algebraic Substitution].

• Given $x$-coordinate (abscissa): $x = 0$
• Given $y$-coordinate (ordinate): $y = 2$

Step 2: Substitution into the Left-Hand Side (LHS)

We isolate the Left-Hand Side (LHS) of the given equation and substitute the given coordinates:

$\text{LHS} = x - 2y$

Substituting $x = 0$ and $y = 2$:

$\text{LHS} = (0) - 2(2)$

$\text{LHS} = 0 - 4$

$\text{LHS} = -4$

Step 3: Comparison with the Right-Hand Side (RHS)

The Right-Hand Side (RHS) of the original equation is a constant:

$\text{RHS} = 4$

Comparing the evaluated LHS with the RHS:

$-4 \neq 4$

$\text{LHS} \neq \text{RHS}$

Because the substitution yields an inequality, the ordered pair $(0, 2)$ does not satisfy the equation.

Step 4: Geometric Interpretation & Verification

Geometrically, a linear equation in two variables represents a straight line on a Cartesian plane. Every point that lies exactly on this line is a solution to the equation. Since $(0, 2)$ does not satisfy the algebraic equation, the point $(0, 2)$ will not lie on the line $x - 2y = 4$.

To graph the line $x - 2y = 4$, we find its intercepts:

• x-intercept (set $y = 0$): $x - 2(0) = 4 \implies x = 4$. Point: $(4, 0)$
• y-intercept (set $x = 0$): $0 - 2y = 4 \implies y = -2$. Point: $(0, -2)$

The visual representation below proves that the point $(0, 2)$ is spatially disconnected from the solution set (the line).

Final Solution: Since substituting $x = 0$ and $y = 2$ results in $\text{LHS} \neq \text{RHS}$ ($-4 \neq 4$), the point $(0, 2)$ is NOT a solution to the equation $x - 2y = 4$.

Given Variables & Initial Setup

We are given a linear equation in two variables and an ordered pair representing a point on the Cartesian plane. Our objective is to determine mathematically whether the given point lies on the line defined by the equation.

• The Linear Equation: $x - 2y = 4$
• The Ordered Pair (Point): $(4, 0)$

Step 1: Coordinate Identification & Substitution

An ordered pair is written in the format $(x, y)$. [Per the fundamental principles of coordinate geometry, a point is a solution to an equation if and only if substituting its coordinates into the equation results in a true mathematical statement].

From the given point $(4, 0)$, we extract the specific coordinate values:

• Abscissa ($x$-coordinate): $x = 4$
• Ordinate ($y$-coordinate): $y = 0$

We substitute these values into the Left-Hand Side (LHS) of the given equation.

$\text{LHS} = x - 2y$

$\text{LHS} = (4) - 2(0)$

Step 2: Algebraic Evaluation

Applying the standard order of operations (PEMDAS/BODMAS), we first perform the multiplication, followed by the subtraction.

$\text{LHS} = 4 - 0$

$\text{LHS} = 4$

Step 3: Verification of Equality

We now compare the evaluated Left-Hand Side (LHS) with the Right-Hand Side (RHS) of the original equation.

• $\text{LHS} = 4$
• $\text{RHS} = 4$

Since $\text{LHS} = \text{RHS}$, the equation holds true. [By the Axiom of Equality, the substitution satisfies the linear equation, proving that the point geometrically lies exactly on the line].

Graphical Representation & Verification

To rigorously verify this result, we can map the equation $x - 2y = 4$ and the point $(4, 0)$ onto a Cartesian coordinate system. The line intersects the x-axis precisely at the point $(4, 0)$.

Final Solution: The ordered pair $(4, 0)$ satisfies the equation $x - 2y = 4$. Therefore, $(4, 0)$ is a valid solution to the given linear equation.

Given Variables & Initial Setup

We are tasked with determining whether the specific coordinate pair is a valid solution to the given linear equation in two variables.

• The Linear Equation: $x - 2y = 4$
• The Coordinate Pair to Test: $(2, 0)$

[Theoretical Foundation: A coordinate pair $(x, y)$ is considered a solution to a linear equation if and only if substituting the $x$ and $y$ values into the equation yields a true mathematical statement. This requires the Left-Hand Side (LHS) of the equation to perfectly equal the Right-Hand Side (RHS) after evaluation.]

Step 1: Isolate the Left-Hand Side (LHS) and Right-Hand Side (RHS)

From the given equation $x - 2y = 4$, we separate the expression into two distinct sides for independent evaluation:

• LHS: $x - 2y$
• RHS: $4$

Step 2: Substitution of the Coordinate Values

The given point is $(2, 0)$. In the Cartesian coordinate system, an ordered pair is formatted as $(x, y)$. Therefore, we extract the following values:

• $x = 2$
• $y = 0$

We substitute these specific values into the LHS expression:

$\text{LHS} = (2) - 2(0)$

Step 3: Algebraic Evaluation

Perform the arithmetic operations according to the standard order of operations (PEMDAS/BODMAS):

$\text{LHS} = 2 - 0$

$\text{LHS} = 2$

Step 4: Comparison and Logical Conclusion

Now, we compare the evaluated LHS with the constant RHS:

• $\text{LHS} = 2$
• $\text{RHS} = 4$

Since $2 \neq 4$, we conclude that $\text{LHS} \neq \text{RHS}$.

[Geometrical Interpretation: Because the equation is not satisfied, the point $(2, 0)$ does not lie on the straight line represented by the equation $x - 2y = 4$.]

Visual Verification: Graphical Representation

Below is the precise Cartesian plot of the line $x - 2y = 4$ alongside the tested point $(2, 0)$. Notice that the line intersects the x-axis at $(4, 0)$, leaving the point $(2, 0)$ distinctly off the line.

Final Solution: Since the substitution of $x = 2$ and $y = 0$ results in $2 \neq 4$, the coordinate pair $(2, 0)$ is NOT a solution to the equation $x - 2y = 4$.

Step 1: Variable Assignment & Initial Setup

To construct a mathematical model of the given real-world scenario, we first define the unknown quantities using algebraic variables. The problem explicitly provides the variable assignments:

• Let the cost of a single notebook be represented by the variable $x$.
• Let the cost of a single pen be represented by the variable $y$.

[These variables represent continuous, non-negative rational numbers, as they denote monetary cost: $x, y \ge 0$].

Step 2: Mathematical Translation of the Statement

The core logical condition provided is: "The cost of a notebook is twice the cost of a pen."

We break down this statement into its algebraic equivalents:

• "The cost of a notebook" translates to $x$.
• "is" translates to the equality operator $=$.
• "twice the cost of a pen" translates to the scalar multiplication $2 \times y$, or simply $2y$.

Step 3: Formulation of the Linear Equation

Equating the translated expressions yields the direct algebraic relationship:

$x = 2y$

[This equation demonstrates a direct proportional relationship between the cost of the notebook and the cost of the pen, where the constant of proportionality is $2$].

Step 4: Standardization of the Equation

The standard form of a linear equation in two variables is expressed as:

$ax + by + c = 0$

where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero ($a^2 + b^2 \neq 0$).

To convert our formulated equation $x = 2y$ into the standard form, we apply the Subtraction Property of Equality by subtracting $2y$ from both sides of the equation:

$x - 2y = 2y - 2y$

$x - 2y = 0$

Here, the coefficients are $a = 1$, $b = -2$, and the constant term is $c = 0$.

Graphical Representation (Visual Verification)

The linear equation $x - 2y = 0$ can be graphed on a Cartesian coordinate system. Because the variables represent cost, the relevant domain and range are restricted to the first quadrant ($x \ge 0, y \ge 0$). The line passes through the origin $(0,0)$, confirming that if a pen costs $0$, a notebook also costs $0$.

Final Solution: The linear equation in two variables representing the given statement is $x - 2y = 0$.

Given Variables & Initial Setup

We are tasked with determining whether the coordinate point $(1, 1)$ is a valid solution to the given linear equation in two variables. The foundational elements of our analysis are:

• The Linear Equation: $x - 2y = 4$
• The Test Coordinate: $(1, 1)$

Step 1: Deconstructing the Coordinate Pair

In the Cartesian coordinate system, an ordered pair is universally represented in the format $(x, y)$. [By the fundamental axiom of coordinate geometry], the first value corresponds to the abscissa (the $x$-coordinate) and the second value corresponds to the ordinate (the $y$-coordinate).

Extracting the values from our test coordinate $(1, 1)$:

• $x = 1$
• $y = 1$

Step 2: Substitution into the Left-Hand Side (LHS)

To verify if the point lies on the line defined by the equation, we must substitute the extracted $x$ and $y$ values into the Left-Hand Side (LHS) of the equation. [Per the algebraic property of substitution, a point is a solution if and only if it satisfies the equality].

The LHS of the equation is given by:

$\text{LHS} = x - 2y$

Substituting $x = 1$ and $y = 1$:

$\text{LHS} = (1) - 2(1)$

$\text{LHS} = 1 - 2$

$\text{LHS} = -1$

Step 3: Comparative Analysis of LHS and RHS

We now compare our evaluated LHS with the Right-Hand Side (RHS) of the original equation.

• $\text{LHS} = -1$
• $\text{RHS} = 4$

Comparing the two values:

$-1 \neq 4$

Since the Left-Hand Side does not equal the Right-Hand Side ($\text{LHS} \neq \text{RHS}$), the coordinate pair $(1, 1)$ fails to satisfy the linear equation.

Step 4: Graphical Verification

To provide rigorous geometric proof, we can visualize the linear equation $x - 2y = 4$ and the point $(1, 1)$ on a Cartesian plane. A point is a solution to an equation if and only if it lies exactly on the line representing that equation.

As demonstrated in the high-precision graph above, the blue line represents the infinite set of solutions for $x - 2y = 4$. The red point representing the coordinate $(1, 1)$ clearly lies outside the trajectory of the line, confirming our algebraic derivation geometrically.

Final Conclusion

Because the substitution of $x = 1$ and $y = 1$ results in a mathematical contradiction ($-1 \neq 4$), the point does not satisfy the equation.

Final Solution: The coordinate $(1, 1)$ is not a solution to the equation $x - 2y = 4$.

Initial Setup & Theoretical Foundation

The given mathematical statement is the equation $y = 3x + 5$. This is a linear equation in two variables, $x$ and $y$. The standard form of a linear equation in two variables is $ax + by + c = 0$. By rearranging the given equation, we obtain $-3x + y - 5 = 0$, where the coefficients are $a = -3$, $b = 1$, and the constant term is $c = -5$.

The standard options for this classical problem are:

• (i) A unique solution
• (ii) Only two solutions
• (iii) Infinitely many solutions

Step 1: Algebraic Analysis of Solutions

A "solution" to a linear equation in two variables is defined as an ordered pair $(x, y)$ that satisfies the equation. In the equation $y = 3x + 5$, $x$ acts as the independent variable and $y$ acts as the dependent variable.

[Per the fundamental properties of real numbers], for every distinct real value assigned to $x$, there exists a corresponding unique real value for $y$. Let us systematically substitute arbitrary real numbers for $x$ to observe the resulting values of $y$:

Because the set of real numbers ($\mathbb{R}$) is infinite, we can choose infinitely many values for $x$. Consequently, we will generate infinitely many corresponding values for $y$. Therefore, the equation possesses infinitely many solutions.

Step 2: Geometric Interpretation

[By the geometric definition of linear equations], any linear equation in two variables represents a straight line on a two-dimensional Cartesian coordinate system. Every single point that lies on this line represents a valid solution $(x, y)$ to the equation.

A straight line extends indefinitely in both directions. Since a line is composed of an infinite number of points, the equation representing that line must inherently have an infinite number of solutions.

Step 3: Visual Proof via Cartesian Mapping

Below is the precise geometric representation of the equation $y = 3x + 5$. The red coordinates highlight specific solutions, demonstrating that the line passes through an infinite continuum of points.

Logical Deduction

Based on both the algebraic substitution method and the geometric mapping, it is evident that the relationship $y = 3x + 5$ is not restricted to one or two specific coordinate pairs. The equation acts as a continuous function mapping the domain of all real numbers to a range of all real numbers.

Final Solution: The true option is "Infinitely many solutions". This is because for every real value of $x$, there is a corresponding unique real value of $y$, and geometrically, the equation represents a straight line composed of an infinite number of points.

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 $x - 2y = 4$.

Initial Setup & Algebraic Isolation

We are given the linear equation in two variables:

$2x + y = 7$

To systematically find solutions for this equation, it is mathematically highly efficient to express one variable in terms of the other. We will isolate the dependent variable $y$ on the left side of the equation [Per the fundamental properties of equality, by subtracting $2x$ from both sides].

$y = 7 - 2x$

A "solution" to a linear equation in two variables is an ordered pair $(x, y)$ that satisfies the equation. Since this is a linear equation defined over the set of real numbers ($\mathbb{R}$), it possesses infinitely many solutions. We will select four arbitrary integer values for the independent variable $x$ to compute the corresponding values of $y$.

Step 1: Calculating the First Solution

Let us assign the value $x = 0$. Substituting this into our isolated equation:

$y = 7 - 2(0)$

$y = 7 - 0$

$y = 7$

This yields our first ordered pair: $(0, 7)$.

Step 2: Calculating the Second Solution

Let us assign the value $x = 1$. Substituting this into the equation:

$y = 7 - 2(1)$

$y = 7 - 2$

$y = 5$

This yields our second ordered pair: $(1, 5)$.

Step 3: Calculating the Third Solution

Let us assign the value $x = 2$. Substituting this into the equation:

$y = 7 - 2(2)$

$y = 7 - 4$

$y = 3$

This yields our third ordered pair: $(2, 3)$.

Step 4: Calculating the Fourth Solution

Let us assign the value $x = 3$. Substituting this into the equation:

$y = 7 - 2(3)$

$y = 7 - 6$

$y = 1$

This yields our fourth ordered pair: $(3, 1)$.

Tabular Representation of Solutions

We can organize the derived coordinates into a data table to clearly demonstrate the linear relationship between $x$ and $y$. Notice that for every $1$ unit increase in $x$, $y$ decreases by $2$ units [This represents the slope of the line, $m = -2$].

Graphical Representation (Cartesian Plane)

Plotting these four points on a Cartesian coordinate system verifies that they are collinear, forming the straight line defined by $2x + y = 7$.

Final Solution: Four distinct solutions for the linear equation $2x + y = 7$ are $(0, 7)$, $(1, 5)$, $(2, 3)$, and $(3, 1)$.