Graph the linear equation y = 2z - 3.

Question

given
graft 1 (0, -1) (1, -3)
graft 2 (1, 5) (0, 3)
graft 3 (0,-3) (1 -1)
graft 4 (0,1) (1,4)

Answer 1

To graph the linear equation $y=2z-3$, it is important to note that this equation is in the form of $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept.

Here, the y-intercept ($b$) is -3, which means the line crosses the y-axis at the point (0, -3). The slope ($m$) is 2, which indicates that for every unit increase in $z$, $y$ increases by 2 units.

To find another point, we can take a specific value for $z$ and calculate $y$.

1. If $z=0$: $$y=2(0)-3=-3\text{(Point is (0, -3))}$$

2. If $z=1$: $$y=2(1)-3=2-3=-1\text{(Point is (1, -1))}$$

Now we can plot these points: (0, -3) and (1, -1).

Let's compare these points to the options provided:

• Graph 1: (0, -1), (1, -3) - Does not match.
• Graph 2: (1, 5), (0, 3) - Does not match.
• Graph 3: (0, -3), (1, -1) - Matches. This is the correct graph for $y=2z-3$.
• Graph 4: (0, 1), (1, 4) - Does not match.

Thus, the correct graph for the equation $y=2z-3$ is Graph 3: (0, -3) and (1, -1).