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Chapter 7: Problem 12

Solve each system by graphing. Check the coordinates of the intersection point in both equations. \(\left\\{\begin{array}{l}y=3 x-4 \\ 2 x+y=1\end{array}\right.\)

Short Answer

Expert verified
The coordinates of the intersection point of the two lines we graphed represent the solution to this system of equations. We verified this solution by substitifying these coordinates into both original equations and confirming that they are true.

Step by step solution

01

Graph the first equation

Start by graphing the equation \(y = 3x - 4\). This is a straight line with a slope of 3 and a y-intercept at -4. Plot a point at -4 on the y-axis which is our y-intercept. From here, move 3 units up and one unit over in the right direction, which yields another point on the line. Plot as many points as you want, and then connect them to create the line graph.
02

Graph the second equation

The second equation is \(2x + y = 1\). To make the graphing easier, we need to rearrange this equation into slope-intercept form, which yields \(y=-2x+1\). Just like in Step 1, we plot a point at 1 on the y-axis, and from there, move 2 units down and one unit over in the right direction for each subsequent point. Connect the plotted points to form the second line.
03

Identify the intersection point

The intersection point of the two lines represents the solution to the system of equations. By looking at the lines on the graph, it's obvious where they intersect. Identify that point.
04

Check the solution

Once you get the intersection point, plug those coordinates \(x, y\) back into both original equations to verify that they hold true in both of them. That confirms that the intersection point is a valid solution.

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Most popular questions from this chapter

a. Create a scatter plot for the data in each table. b. Use the shape of the scatter plot to determine if the data are best modeled by a linear function, an exponential function, a logarithmic function, or a quadratic function. $$ \begin{array}{|c|c|} \hline \boldsymbol{x} & \boldsymbol{y} \\ \hline 0 & 4 \\ \hline 1 & 1 \\ \hline 2 & 0 \\ \hline 3 & 1 \\ \hline 4 & 4 \\ \hline \end{array} $$

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