To solve this system of equations by graphing, let's go through each step systematically.

Given system of equations:

1. $y = x - 1$
2. $y = 2x + 1$

Steps:

1. Graph the first equation $y = x - 1$:

   • This is a straight line with a slope of 1 and a y-intercept of $-1$ (the point $(0, -1)$).
   • You can find another point by substituting another $x$ value. For example, if $x = 1$, then $y = 1 - 1 = 0$. So, another point on the line is $(1, 0)$.
   • Plot these points: $(0, -1)$ and $(1, 0)$, and draw a line through them.

2. Graph the second equation $y = 2x + 1$:

   • This line has a slope of 2 and a y-intercept of 1 (the point $(0, 1)$).
   • If $x = 1$, then $y = 2(1) + 1 = 3$, so another point is $(1, 3)$.
   • Plot the points $(0, 1)$ and $(1, 3)$, and draw a line through them.

3. Find the intersection point:

   • The solution to the system is where the two lines intersect.
   • After graphing both lines, observe the point where they cross, which will be the solution to the system of equations.

Now, let's calculate the intersection algebraically:

We can set the two equations equal to each other: $x - 1 = 2x + 1$

Solve for $x$: $x - 2x = 1 + 1$ $-x = 2$ $x = -2$

Substitute $x = -2$ back into one of the original equations (I'll use $y = x - 1$): $y = -2 - 1 = -3$

So, the solution is $(-2, -3)$.

Plot the point $(-2, -3)$ on the graph:

• This is the point where the two lines intersect, which represents the solution to the system of equations.

Would you like me to provide additional details or clarify any steps?

Here are 5 related questions:

1. How do you calculate the slope of a line from its equation?
2. What happens if the two lines in a system of equations are parallel?
3. How can we solve a system of equations using substitution instead of graphing?
4. How do we identify the y-intercept of a line from its equation?
5. What does it mean if the two lines in a system of equations coincide?

Tip: When graphing linear equations, always start by plotting the y-intercept, as it's the easiest point to find from the equation.