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Solve the equation graphically. $$-2 x+3=8 x$$

Short Answer

Expert verified
The solution to the equation \(-2x + 3 = 8x\) is the x-coordinate of the intersection point of the lines represented by \(-2x + 3\) and \(8x\).

Step by step solution

01

Plot the First Equation

The first equation is a linear equation in the form \(y = mx + c\) which graphically represents a straight line. In the case of \(-2x + 3\), the slope \(m\) is -2 and the y-intercept \(c\) is 3. So, set y equal to \(-2x + 3\), choose a range for x and compute corresponding y-values to create the (x,y) pairs. Plot these points on a graph, and connect them to represent the line for the first equation.
02

Plot the Second Equation

The second equation is \(8x\), a line with slope 8 and goes through the origin. Similar to step 1, set y equal to \(8x\), compute the corresponding y-values for a chosen range of x-values and plot these points on the same graph as in Step 1, which will represent the line for the second equation.
03

Find the Intersection Point

Analyze the graph and locate the intersection point of the two lines. The x-coordinate of this point is the solution to the original equation.

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Key Concepts

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

Linear Equations
Linear equations are fundamental building blocks in algebra and geometry. They are called "linear" because their graph is a straight line in a two-dimensional space. A standard form for a linear equation in two variables is:
  • \( y = mx + c \)
Here, "\(m\)" represents the slope of the line, determining how steep the line is; "\(c\)" is the y-intercept, where the line crosses the y-axis.
This representation helps us visualize relationships in a clear manner. Understanding the behavior of linear equations is crucial for solving various mathematical and real-world problems.
In our exercise, both equations given, \(-2x + 3\) and \(8x\), are linear because they match the form \(y = mx + c\). The graph of any linear equation will always be a straight line, defined uniquely by its slope and intercepts.
Solution of Systems of Equations
Solving systems of equations involves finding the set of values that satisfy multiple equations simultaneously. This is commonly encountered when graphing, where the goal is to locate the set of points meeting the conditions of all equations involved.
Graphical solutions are very useful because they provide a visual representation of these simultaneous solutions. In the case of two linear equations, such as in our exercise, the solution is often the point where the lines intersect on a graph.
There are different techniques to solve systems of equations, including:
  • Graphical method
  • Substitution method
  • Elimination method
For our specific problem, we use the graphical method to indicate where the two lines - represented by the original equations - intersect, revealing the solution to the system. The intersection point gives us both a clear visual cue and a numerical solution.
Intersection of Lines
The intersection of lines is a critical concept when dealing with multiple linear equations. It refers to the point where two lines meet or cross each other on a graph. This point represents a solution that satisfies both equations simultaneously.
Understanding intersections involves recognizing how the equations correlate on a graph. For any given pair of lines:
  • If they intersect, there is a single solution at the intersection point.
  • If they are parallel and never intersect, there is no solution.
  • If they coincide (overlap entirely), there are infinitely many solutions.
In our example, the intersection point of \(-2x + 3\) and \(8x\) on a graph reveals the solution to the original equation. This solution gives us the exact x-coordinate where both equations have matching y-values, providing a tangible answer to the problem.

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Most popular questions from this chapter

In your own words, explain how to fit a model to a set of data using a graphing utility.

Use the zero or root feature of a graphing utility to approximate the solution of the logarithmic equation. $$\log _{10} x+e^{0.5 x}=6$$

Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$2 x^{2} e^{2 x}+2 x e^{2 x}=0$$

Solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer(s) using a graphing utility. $$\log _{10} 8 x-\log _{10}(1+\sqrt{x})=2$$

A beaker of liquid at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C}\) The temperature of the liquid is measured every 5 minutes for a period of \(\frac{1}{2}\) hour. The results are recorded in the table, where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$\begin{array}{|c|c|}\hline \text { Time, } t 0 & \text { Temperature, } T\\\\\hline 0 & 78.0^{\circ} \\\5 & 66.0^{\circ} \\\10 & 57.5^{\circ} \\\15 & 51.2^{\circ} \\\20 & 46.3^{\circ} \\\25 & 42.5^{\circ} \\\30 & 39.6^{\circ} \\\\\hline\end{array}$$ (a) Use the regression feature of a graphing utility to find a linear model for the data. Use the graphing utility to plot the data and graph the model in the same viewing window. Do the data appear linear? Explain. (b) Use the regression feature of the graphing utility to find a quadratic model for the data. Use the graphing utility to plot the data and graph the model in the same viewing window. Do the data appear quadratic? Even though the quadratic model appears to be a good fit, explain why it might not be a good model for predicting the temperature of the liquid when \(t=60.\) (c) The graph of the temperature of the room should be an asymptote of the graph of the model. Subtract the room temperature from each of the temperatures in the table. Use the regression feature of the graphing utility to find an exponential model for the revised data. Add the room temperature to this model. Use the graphing utility to plot the original data and graph the model in the same viewing window. (d) Explain why the procedure in part (c) was necessary for finding the exponential model.

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