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Graph each linear system, either by hand or using a graphing device. Use the graph to determine whether the system has one solution, no solution, or infinitely many solutions. If there is exactly one solution, use the graph to find it. $$\left\\{\begin{aligned}12 x+15 y &=-18 \\\2 x+\frac{5}{2} y &=-3\end{aligned}\right.$$

Short Answer

Expert verified
The system has infinitely many solutions, as both equations represent the same line.

Step by step solution

01

Convert to Slope-Intercept Form

To graph each equation, we first convert them into the slope-intercept form, which is \(y = mx + b\). This makes it easier to graph.For the first equation: \[12x + 15y = -18\]Divide every term by 15:\[y = -\frac{4}{5}x - \frac{6}{5}\]For the second equation:\[2x + \frac{5}{2}y = -3\]Multiply every term by 2 to eliminate the fraction:\[4x + 5y = -6\]Now, solve for \(y\):\[y = -\frac{4}{5}x - \frac{6}{5}\]
02

Analyze the Equations

Look at both equations we have now:- First equation: \(y = -\frac{4}{5}x - \frac{6}{5}\)- Second equation: \(y = -\frac{4}{5}x - \frac{6}{5}\)Notice that the two equations are identical, meaning they will produce the same line when graphed.
03

Graph the Equations

Since both equations are identical, graph just one of them.The line has a slope of \(-\frac{4}{5}\) and a y-intercept of \(-\frac{6}{5}\). Start by plotting the y-intercept at \((0, -\frac{6}{5})\). Then use the slope to find another point. From this intercept, go down 4 units and to the right 5 units to plot another point.Draw a line through these points.
04

Determine the Type of Solution

Since both equations represent the same line, the system of equations has infinitely many solutions. This happens because every point on the line is a solution that satisfies both equations.

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form is a way of expressing the equation of a line. It is written as \(y = mx + b\). This form is very useful when you want to graph a linear equation. Here, "\(m\)" represents the slope of the line, which describes how steep the line is.
It tells us how much the line goes up or down for each step it moves horizontally. The "\(b\)" is the y-intercept, which is the point where the line crosses the y-axis. It tells us where the line starts on the y-axis when \(x = 0\).

Converting an equation to this form helps in easily identifying the slope and y-intercept, making graphing straightforward. Let's see why:
  • Start with an equation in another form, like standard form \(Ax + By = C\).
  • Manipulate it to solve for \(y\) in terms of \(x\).
  • Ensure the equation looks like \(y = mx + b\).
With the slope and intercept identified, you can graph the line by first plotting the y-intercept, then using the slope to find additional points. This simple form makes graphing less daunting.
Identical Equations
When we say two equations are identical, it means they are the exact same line when graphed.
This occurs because their slope and intercept values are the same. In the example given, both equations describe the same line \(y = -\frac{4}{5}x - \frac{6}{5}\).
This implies that no matter what \(x\) value you choose, the value of \(y\) computed from both equations will be the same.

Identical equations are an indication that two seemingly different lines in a system of linear equations are, in fact, the same.
  • They share all their points.
  • Their slopes \(m\) are equal.
  • Their y-intercepts \(b\) are equal.
This can happen after simplifying or transforming the equations through equivalent steps, like factoring or eliminating fractions.
Thus, if you find that two equations are identical when in slope-intercept form, they will graph as one line.
Infinitely Many Solutions
The phrase "infinitely many solutions" occurs when every point on one line is also a point on the other line in a system of linear equations.
As seen in the step-by-step solution, once the equations are determined to be identical, the graphing of these presents only one line.

This actually means that each point on this line satisfies both equations simultaneously.
  • Both equations describe the same linear path.
  • Any point on this line will work in both equations.
Therefore, there is not just one solution, or none at all, but an infinite number. It is important to recognize this scenario as it fundamentally means the relationship between the variables described by one line is identically described by the other.
Seeing a single line for two equations when graphed confirms their equivalency, resulting in infinitely many solutions.

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Most popular questions from this chapter

Use the definition of determinant and the elementary row and column operations to explain why matrices of the following types have determinant 0. (a) A matrix with a row or column consisting entirely of zeros (b) A matrix with two rows the same or two columns the same (c) A matrix in which one row is a multiple of another row, or one column is a multiple of another column

A man and his daughter manufacture unfinished tables and chairs. Each table requires 3 hours of sawing and 1 hour of assembly. Each chair requires 2 hours of sawing and 2 hours of assembly. Between the two of them, they can put in up to 12 hours of sawing and 8 hours of assembly work each day. Find a system of inequalities that describes all possible combinations of tables and chairs that they can make daily. Graph the solution set.

We have used the Zero-Product Property to solve algebraic equations. Matrices do not have this property. Let \(O\) represent the \(2 \times 2\) zero matrix $$O=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$ Find \(2 \times 2\) matrices \(A \neq O\) and \(B \neq O\) such that \(A B=O\) Can you find a matrix \(A \neq O\) such that \(A^{2}=O ?\)

Find the inverse of the matrix if it exists. $$\left[\begin{array}{rrr} 3 & -2 & 0 \\ 5 & 1 & 1 \\ 2 & -2 & 0 \end{array}\right]$$

Classroom Use A small school has 100 students who occupy three classrooms: \(A, B,\) and \(C\). After the first period of the school day, half the students in room A move to room B one-fifth of the students in room B move to room \(\mathrm{C}\), and one-third of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room?

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