In the image, the problem involves two linear equations:

1. $y = 2x + 3$
2. $y = -x + 1$

We need to do the following:

a) Find the intersection point of the lines.

To find the intersection, we solve the system of equations. Since both are equal to $y$, we set the right-hand sides equal to each other:

$2x + 3 = -x + 1$

Now, solve for $x$:

$2x + x = 1 - 3$ $3x = -2$ $x = -\frac{2}{3}$

Now, substitute $x = -\frac{2}{3}$ into either of the original equations to find $y$. Using $y = 2x + 3$:

$y = 2\left(-\frac{2}{3}\right) + 3 = -\frac{4}{3} + 3 = -\frac{4}{3} + \frac{9}{3} = \frac{5}{3}$

Thus, the intersection point is $\left( -\frac{2}{3}, \frac{5}{3} \right)$.

b) Graph the lines and show the intersection point.

To graph the lines:

• The first equation $y = 2x + 3$ has a slope of 2 and a y-intercept of 3.
• The second equation $y = -x + 1$ has a slope of -1 and a y-intercept of 1.

The lines intersect at $\left( -\frac{2}{3}, \frac{5}{3} \right)$, which can be verified on the graph.

Would you like me to generate a graph to visualize the intersection point?

Related Questions:

1. What is the slope of each line?
2. How can we verify the solution by substituting the values back into both equations?
3. How do we interpret the intersection point geometrically?
4. What is the general method for solving two linear equations?
5. How does the slope affect the steepness of the line?

Tip:

When solving systems of linear equations, substitution or elimination methods are common techniques. Use substitution when one variable is easily isolated, as in this case!