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Chapter 2: Problem 11

Find the solution set of each equation. $$3 x-6 y=0$$

Short Answer

Expert verified
The solution set is \{(x, \frac{1}{2}x) | x \in \mathbb{R}\}.

Step by step solution

01

Identify the Equation Type

The given equation is in the form \(3x - 6y = 0\). This is a linear equation in two variables, representing a line.
02

Rearrange Equation in Slope-Intercept Form

Rearrange the equation into the slope-intercept form \(y = mx + b\) to find the relationship between \(x\) and \(y\). Start by solving for \(y\):\[3x - 6y = 0\] Add \(6y\) to both sides: \[3x = 6y\] Divide each term by 6 to solve for \(y\):\[y = \frac{1}{2}x\]
03

Express the Solution Set

Since \(y = \frac{1}{2}x\), any point \((x, y)\) that satisfies this relationship is a solution. The solution set is \{(x, \frac{1}{2}x) | x \in \mathbb{R}\}.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solution Set
A solution set refers to all the possible solutions that satisfy a given equation. In our exercise, we're dealing with a linear equation. A linear equation typically represents a line in a 2D plane.

For the equation \(3x - 6y = 0\), we found that the solution must satisfy the relationship \(y = \frac{1}{2}x\). This means that for any real number \(x\), there will be a corresponding \(y\) such that together, they form a point on the line. Therefore, the solution set can be described as:
  • All points \((x, y)\) where \(y = \frac{1}{2}x\)
  • \(x\) can be any real number, \(x \in \mathbb{R}\)
In simpler terms, if you choose any number for \(x\) and multiply it by \(\frac{1}{2}\), the resulting \(y\) gives you a point on the line that the equation describes.
Slope-Intercept Form
The slope-intercept form of a linear equation is particularly useful for quickly understanding the slope and y-intercept of the line. This form is written as \(y = mx + b\), where:
  • \(m\) is the slope of the line.
  • \(b\) is the y-intercept, the point where the line crosses the y-axis.
In our exercise, by rearranging \(3x - 6y = 0\) into slope-intercept form, we found it to be \(y = \frac{1}{2}x\).

Here, \(m = \frac{1}{2}\) indicates that for every 1 unit increase in \(x\), \(y\) increases by \(0.5\) units. Notice that there is no \(b\), meaning the y-intercept is \(0\). Thus, this line passes through the origin \((0,0)\).

The slope-intercept form helps us easily graph the line and better understand its orientation and position on the Cartesian plane.
Two Variables
Linear equations such as the one in our exercise involve two variables, usually denoted as \(x\) and \(y\). Each variable represents a dimension in a 2D space.
  • \(x\) is often referred to as the independent variable.
  • \(y\) is commonly called the dependent variable, as its value depends on \(x\).
In the equation \(3x - 6y = 0\), both \(x\) and \(y\) are essential for forming a complete description of the line. Changing \(x\) results in a change in \(y\), consistent with the given relationship \(y = \frac{1}{2}x\).

By understanding this dynamic between the two variables, we can see why the solution set forms a line: for each \(x\), only one \(y\) satisfies the relationship, perfectly plotting a line with an infinite number of solutions.

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Most popular questions from this chapter

Show that if \(a d-b c \neq 0\), then the system $$\begin{array}{l} a x+b y=r \\ c x+d y=s \end{array}$$ has a unique solution.

Recall that the cross product of vectors u and v is a vector \(\mathbf{u} \times \mathbf{v}\) that is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). ( See Exploration: The Cross Product in Chapter 1.) If$$\mathbf{u}=\left[\begin{array}{l} u_{1} \\ u_{2} \\ u_{3} \end{array}\right] \quad \text { and } \quad \mathbf{v}=\left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right]$$ show that there are infinitely many vectors $$\mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]$$ that simultaneously satisfy \(\mathbf{u} \cdot \mathbf{x}=0\) and \(\mathbf{v} \cdot \mathbf{x}=0\) and that all are multiples of $$\mathbf{u} \times \mathbf{v}=\left[\begin{array}{l} u_{2} v_{3}-u_{3} v_{2} \\ u_{3} v_{1}-u_{1} v_{3} \\ u_{1} v_{2}-u_{2} v_{1} \end{array}\right]$$

Determine whether the lines \(x=\) \(\mathbf{p}+\) su and \(\mathbf{x}=\mathbf{q}+\) tv intersect and, if they do, find their point of intersection. $$\mathbf{p}=\left[\begin{array}{l} 3 \\ 1 \\ 0 \end{array}\right], \mathbf{q}=\left[\begin{array}{r} -1 \\ 1 \\ -1 \end{array}\right], \mathbf{u}=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right], \mathbf{v}=\left[\begin{array}{l} 2 \\ 3 \\ 1 \end{array}\right]$$

determine if the sets of vectors in Exercises \(22-31\) are linearly in dependent. If for any of these, the answer can be determined by inspection (i.e., without calculation), state why. For any sets that are linearly dependent, find a dependence relationship among the vectors. $$\left[\begin{array}{l} 0 \\ 1 \\ 2 \end{array}\right],\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right],\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right]$$

For what value(s) of \(k\), if any, will the systems have (a) no solution, (b) a unique solution, and (c) infinitely many solutions? $$\begin{array}{l} k x+2 y=3 \\ 2 x-4 y=-6 \end{array}$$
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