$7$ . Draw the graphs of the equations $x-y+1=0$ and $3 x+2 y-12=0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis, and shade the triangular region.

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7

. Draw the graphs of the equations

x-y+1=0

and

3 x+2 y-12=0

. Determine the coordinates of the vertices of the triangle formed by these lines and the

x

-axis, and shade the triangular region.

Rearrange equation to slope-intercept form: First, let's rearrange the equation $x - y + 1 = 0$ to slope-intercept form $(y = mx + b)$ to find where it crosses the y-axis. $\newline$ $y = x + 1$
Find x-intercepts of both lines: Now, let's do the same for the equation $3x + 2y - 12 = 0$ . $\newline$ $2y = -3x + 12$ $\newline$ $y = -\frac{3}{2}x + 6$
Find point of intersection: Next, we'll find the $x$ -intercepts of both lines by setting $y$ to $0$ and solving for $x$ . For the first line, $x + 1 = 0$ , so $x = -1$ . For the second line, $-\frac{3}{2}x + 6 = 0$ , so $x = 4$ .
Substitute and solve system of equations: Now we need to find the point of intersection of the two lines by solving the system of equations: $\newline$ $x - y + 1 = 0$ $\newline$ $3x + 2y - 12 = 0$
Find intersection point: Let's use substitution or elimination. I'll use substitution. From the first equation, $y = x + 1$ . Substitute this into the second equation: $\newline$ $3x + 2(x + 1) - 12 = 0$ $\newline$ $3x + 2x + 2 - 12 = 0$ $\newline$ $5x - 10 = 0$ $\newline$ $x = $2$
Identify vertices of the triangle: Now substitute $x = 2$ into $y = x + 1$ to find $y$:\[y = 2 + 1\]\[y = 3\]So the point of intersection is $(2, 3)$.
Shade triangular region: The vertices of the triangle are the x-intercepts of both lines and their point of intersection.$\newline$First vertex (x-intercept of the first line): $(-1, 0)$$\newline$Second vertex (x-intercept of the second line): $(4, 0)$$\newline$Third vertex (intersection point): $(2, 3)$
Shade triangular region: The vertices of the triangle are the x-intercepts of both lines and their point of intersection.$\newline$First vertex (x-intercept of the first line): $(-1, 0)$$\newline$Second vertex (x-intercept of the second line): $(4, 0)$$\newline$Third vertex (intersection point): $(2, 3)$ Finally, we shade the triangular region formed by these three vertices and the lines connecting them.