Question

graph this system of equations and determine the number of solutions.
$y = \frac{2}{3}x - 2$
$y = x - 1$
click to select points on the graph.

Explanation:

Step1: Analyze the first equation

The equation \( y = \frac{2}{3}x - 2 \) is in slope - intercept form (\(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept). The slope \(m_1=\frac{2}{3}\) and the y - intercept \(b_1=- 2\). To graph this line, we can start by plotting the y - intercept \((0,-2)\). Then, using the slope (rise over run), from \((0,-2)\), we rise 2 units and run 3 units to get another point, for example, \((3,0)\) (since \(-2 + 2=0\) and \(0 + 3 = 3\)).

Step2: Analyze the second equation

The equation \( y=x - 1\) is also in slope - intercept form. The slope \(m_2 = 1\) and the y - intercept \(b_2=-1\). We plot the y - intercept \((0,-1)\). Then, using the slope (rise 1, run 1), from \((0,-1)\), we can get another point \((1,0)\) (since \(-1+1 = 0\) and \(0 + 1=1\)).

Step3: Determine the number of solutions

Since the two lines have different slopes (\(m_1=\frac{2}{3}\) and \(m_2 = 1\)), they are not parallel and will intersect at exactly one point. To find the intersection point algebraically (to confirm), we set the two equations equal to each other:
\(\frac{2}{3}x-2=x - 1\)
Subtract \(\frac{2}{3}x\) from both sides: \(-2=\frac{1}{3}x-1\)
Add 1 to both sides: \(-1=\frac{1}{3}x\)
Multiply both sides by 3: \(x=-3\)
Substitute \(x = - 3\) into \(y=x - 1\), we get \(y=-3 - 1=-4\). So the lines intersect at \((-3,-4)\), meaning there is 1 solution.

Answer:

The system of equations has 1 solution.