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Chapter 2: Problem 18

Draw graphs corresponding to the given linear systems. Determine geometrically whether each system has a unique solution, infinitely many solutions, or no solution. Then solve each system algebraically to confirm your answer. $$\begin{array}{rr}0.10 x-0.05 y= & 0.20 \\\\-0.06 x+0.03 y= & -0.12\end{array}$$

Short Answer

Expert verified
Infinitely many solutions; both equations represent the same line.

Step by step solution

01

Transform the equations

First, let's convert the given equations into a more recognizable form. We have two equations:1. \(0.10x - 0.05y = 0.20\)2. \(-0.06x + 0.03y = -0.12\)Multiply the entire first equation by 10 to remove decimals:\(x - 0.5y = 2\)Multiply the entire second equation by 50 to remove decimals:\(-3x + 1.5y = -6\)
02

Rewrite in slope-intercept form

Rewrite each equation in slope-intercept form (\(y = mx + b\)). For the first equation:\[x - 0.5y = 2\]Solve for \(y\):\(0.5y = x - 2\)\(y = 2x - 4\)For the second equation:\[-3x + 1.5y = -6\]Solve for \(y\):\(1.5y = 3x - 6\)\(y = 2x - 4\)
03

Graph the equations

Both equations are now \(y = 2x - 4\), which means they are the same line. To graph, plot the y-intercept (-4) on the y-axis and use the slope (2) to find other points on the line.
04

Geometrical Interpretation

Since both equations represent the same line, the system has **infinitely many solutions**. Graphically, this means all points on the line are solutions.
05

Solve algebraically to confirm

Algebraically, we substitute the equation back into the system. Given both equations are identical:Substitute one equation into the other:\(y = 2x - 4\)Substitute back, both equations are true:\(y = 2x - 4\)This confirms that they coincide over all values that satisfy \(y = 2x - 4\). Hence, **infinitely many solutions**.

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Key Concepts

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

Graphs of Linear Equations
Graphs of linear equations involve plotting straight lines on a coordinate plane. Each equation corresponds to a line. When graphing, it's crucial to transform each equation into the slope-intercept form, which is expressed as \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
For the equations given in the exercise, once in slope-intercept form, you see they are indeed identical: \(y = 2x - 4\). This means these equations graph onto the same line.
  • The y-intercept (-4) tells us where the line crosses the y-axis.
  • The slope (2) defines the steepness and direction of the line.
  • To plot, start at -4 on the y-axis, and for every unit increase in \(x\), \(y\) increases by 2.
Unique Solution
A unique solution in a linear system refers to a single intersection point of two lines on a graph. In the context of linear equations, a unique solution implies the two lines cross at one point, meaning they have different slopes and are not parallel.

When looking at the graphs for systems with unique solutions:
  • The slopes of the lines will not be the same, ensuring the lines will eventually intersect.
  • The intersection point gives the values for \(x\) and \(y\) that solve both equations simultaneously.
  • It is important to check that both equations are not multiples of each other, which guarantees they are indeed different lines.
In our given system, since the equations are identical, we don't have a unique solution but rather infinite ones. Understanding this concept helps differentiate between varied solution scenarios.
Infinitely Many Solutions
In a linear system, having infinitely many solutions means that the equations describe the same line. As a result, every point on this line satisfies both equations. The steps to arrive at this conclusion include:
  • Transform each equation to the slope-intercept form.
  • If both equations simplify to the same linear equation, the lines completely overlap.
  • Graphically, you will only see one line rather than two separate ones.
The exercise showed that both equations simplify to \(y = 2x - 4\). Hence,
  • All solutions are points on this line;
  • Any \(x\) you choose will correspond to a specific \(y\) point following the equation \(y = 2x - 4\).
Geometric Interpretation
Geometric interpretation is a visual method to understand solutions of linear systems. It involves analyzing the graph of the equations:
  • A unique solution appears as an intersection point of two lines.
  • Infinitely many solutions occur when the lines coincide completely, i.e., all points on one line are also on the other.
  • No solution is present when the lines are parallel but not the same, implying they never meet.
In our exercise, the identical graphs suggested an infinite number of solutions, since the entire line of \(y = 2x - 4\) is shared by both equations. Using graphical interpretation allows us to quickly determine the nature of the solution.
Algebraic Solution
An algebraic solution involves solving the system of equations mathematically to confirm the graphical findings. Here's how you do it:
  • Take each transformed equation and equate them.
  • Substitute back into one of the original equations if needed.
  • If the statements you derive are true for any \(x\) or \(y\), it confirms overlapping lines, hence, infinite solutions.
In the exercise, both equations were identical to \(y = 2x - 4\) after transformation. Thus, they algebraically confirmed that any point satisfying one equation satisfies the other, reaffirming infinite solutions.

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Most popular questions from this chapter

(a) Suppose that vector \(\mathbf{w}\) is a linear combination of vectors \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}\) and that each \(\mathbf{u}_{i}\) is a linear combination of vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{m} .\) Prove that \(\mathbf{w}\) is a linear combination of \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\) and therefore \(\operatorname{span}\left(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}\right) \subseteq \operatorname{span}\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\right)\) (b) In part (a), suppose in addition that each \(\mathbf{v}_{j}\) is also a linear combination of \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}\). Prove that \(\operatorname{span}\left(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}\right)=\operatorname{span}\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\right)\) (c) Use the result of part (b) to prove that \\[ \mathbb{R}^{3}=\operatorname{span}\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]\right) \\] \(\left[\text {Hint: We know that } \mathbb{R}^{3}=\operatorname{span}\left(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right) .\right]\)

Determine by inspection (i.e., without performing any calculations) whether a linear system with the given augmented matrix has a unique solution, infinitely many solutions, or no solution. Justify your answers. \(\left[\begin{array}{lllll|l}1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 4 & 3 & 2 & 1 \\\ 7 & 7 & 7 & 7 & 7 & 7\end{array}\right]\)

Compute the first four iterates, using the zero vector as the initial approximation, to show that the Gauss-Seidel method diverges. Then show that the equations can be rearranged to give a strictly diagonally dominant coefficient matrix, and apply the Gauss-Seidel method to obtain an approximate solution that is accurate to within 0.001. $$\begin{aligned}x_{1}-4 x_{2}+2 x_{3} &=2 \\\2 x_{2}+4 x_{3} &=1 \\\6 x_{1}-x_{2}-2 x_{3} &=1\end{aligned}$$

Set up and solve an appropriate system of linear equations to answer the questions. From elementary geometry we know that there is a unique straight line through any two points in a plane. Less well known is the fact that there is a unique parabola through any three noncollinear points in a plane. For each set of points below, find a parabola with an equation of the form \(y=a x^{2}+\) \(b x+c\) that passes through the given points. (Sketch the resulting parabola to check the validity of your answer. (a) \((0,1),(-1,4),\) and (2,1) (b) \((-3,1),(-2,2),\) and (-1,5)

Find a system of linear equations that has the given matrix as its augmented matrix. $$\left[\begin{array}{rrr|r}0 & 1 & 1 & 1 \\\1 & -1 & 0 & 1 \\\2 & -1 & 1 & 1\end{array}\right]$$
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