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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. $$\begin{array}{r} x-2 y=7 \\ 3 x+y=7 \end{array}$$

Short Answer

Expert verified
Unique solution at the intersection point \((3, -2)\).

Step by step solution

01

Identify the Lines

The given system of equations is \(x - 2y = 7\) and \(3x + y = 7\). Each equation represents a line in the coordinate plane. We need to express each equation in the form \(y = mx + b\) to identify the slope and y-intercept.
02

Convert Equations to Slope-Intercept Form

For the first equation \(x - 2y = 7\), solve for \(y\):\[x - 2y = 7 \-2y = -x + 7 \y = \frac{1}{2}x - \frac{7}{2}\]For the second equation \(3x + y = 7\), solve for \(y\):\[3x + y = 7 \y = -3x + 7\]

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Key Concepts

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

Graphing Linear Equations
Graphing linear equations involves plotting the solutions of a linear equation on a coordinate plane. Each linear equation represents a straight line. These lines can either intersect, be parallel, or coincide with each other based on their slopes and intercepts. To graph a linear equation, you should have at least two points that satisfy the equation or use its slope and y-intercept.
You can start by converting the equation into a more usable form, like the slope-intercept form, which makes it easy to identify the main characteristics of the line. Let's apply this to our equations: x - 2y = 7 and 3x + y = 7. We transform them into y = \(\frac{1}{2}x - \frac{7}{2}\) and y = -3x + 7, respectively.
Once transformed, you can plot them easily on a graph by using the y-intercept as a starting point, and using the slope to find other points. This helps determine if the lines intersect (unique solution), are parallel (no solution), or overlap completely (infinite solutions).
Slope-Intercept Form
The slope-intercept form of a linear equation is given by y = mx + b. In this format, m represents the slope of the line, and b refers to the y-intercept, the point where the line crosses the y-axis. This form is particularly useful because it provides direct insight into how the line behaves and how it can be easily graphed.
  • Slope (\(m\)): Represents the steepness of the line and the direction in which it moves. If the slope is positive, the line ascends from left to right; if negative, it descends.
  • Y-intercept (\(b\)): The value of y when x is zero, making it a critical starting point for plotting the graph.
Applying it to our system of equations:
  • The first equation transforms to \(y = \frac{1}{2}x - \frac{7}{2}\), with a slope of \(\frac{1}{2}\) and y-intercept of \(-\frac{7}{2}\).
  • The second becomes \(y = -3x + 7\), with a slope of \(-3\) and y-intercept of \(7\).
This simple transformation helps us analyze how the lines will interact on a graph.
Unique Solution Determination
Determining the number of solutions a system of linear equations has involves analyzing their graphical representation. In a coordinate system, intersecting lines indicate a unique solution where the lines cross.
Parallel lines have no points of intersection, indicating no solution. If the lines overlap completely, sharing all points along their length, there are infinitely many solutions.
With our equations already in the slope-intercept form, identifying intersections becomes straightforward. The first equation \(y = \frac{1}{2}x - \frac{7}{2}\) has a slope of \(\frac{1}{2}\), while the second equation \(y = -3x + 7\) possesses a different slope, \(-3\). Different slopes guarantee they will eventually intersect, hence, the system has a unique solution. Validate this by solving algebraically, confirming both equations have one point in common, ensuring the system has precisely one solution.

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Most popular questions from this chapter

Show that \(\mathbb{R}^{3}=\operatorname{span}\left(\left[\begin{array}{l}1 \\\ 1 \\ 1\end{array}\right],\left[\begin{array}{r}1 \\ 1 \\\ -1\end{array}\right],\left[\begin{array}{r}-1 \\ 1 \\\ 1\end{array}\right]\right)\)

A coffee merchant sells three blends of coffee. A bag of the house blend contains 300 grams of Colombian beans and 200 grams of French roast beans. A bag of the special blend contains 200 grams of Colombian beans, 200 grams of Kenyan beans, and 100 grams of French roast beans. A bag of the gourmet blend contains 100 grams of Colombian beans, 200 grams of Kenyan beans, and 200 grams of French roast beans. The merchant has on hand 30 kilograms of Colombian beans, 15 kilograms of Kenyan beans, and 25 kilograms of French roast beans. If he wishes to use up all of the beans, how many bags of each type of blend can be made?

When \(p\) is not prime, extra care is needed in solving a linear system (or, indeed, any equation) over \(\mathbb{Z}_{p}\) Using Gaussian elimination, solve the following system over \(\mathbb{Z}_{6} .\) What complications arise? \\[\begin{array}{l}2 x+3 y=4 \\\4 x+3 y=2\end{array}\\]

Consider the matrix \(A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right] .\) Show that any of the three types of elementary row operations can be used to create a leading 1 at the top of the first column. Which do you prefer and why?

Show that the given matrices are row equivalent and find a sequence of elementary row operations that will convert A into B. \(A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{rr}3 & -1 \\ 1 & 0\end{array}\right]\)

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