It is necessary to find the coordinates of the vertices of the triangle, each of which is the intersection of two of the three lines.

The intersection of the lines of the equations $\displaystyle x = -4$ and $\displaystyle y = x$ can easily be found by noting that since $\displaystyle x = -4$, by substitution, $\displaystyle y = -4$, making the point of intersection $\displaystyle (-4. -4)$.

The intersection of the lines of the equations $\displaystyle x = -4$ and $\displaystyle y = 4x- 7$ can be found by substituting  $\displaystyle -4$ for $\displaystyle x$ in the latter equation and evaluating $\displaystyle y$:

$\displaystyle y = 4 (-4)- 7 = -16 - 7 = -23$

The point of intersection is $\displaystyle (-4, -23)$.

The intersection of the lines of the equations $\displaystyle y = x$ and $\displaystyle y = 4x- 7$ can be found by substituting  $\displaystyle x$ for $\displaystyle y$ in the latter equation and solving for $\displaystyle x$:

$\displaystyle 4x- 7 = x$

$\displaystyle 4x- 7 - x + 7 = x - x + 7$

$\displaystyle 3x = 7$

$\displaystyle \frac{3x }{3}= \frac{7}{3}$

$\displaystyle x = 2\frac{1}{3}$

$\displaystyle y = x$, so $\displaystyle y = 2\frac{1}{3}$, and this point of intersection is $\displaystyle \left ( 2\frac{1}{3}, 2\frac{1}{3} \right )$.

The lines in question are graphed below, and the triangle they bound is shaded:

We can take the vertical side as the base of the triangle; its length is the difference of the $\displaystyle y$-coordinates:

$\displaystyle b = -4 - (-23) = 19$

The height is the horizontal distance from this side to the opposite side, which is the difference of the $\displaystyle x$-coordinates:

$\displaystyle h= 2 \frac{1}{3} - (-4) = 6 \frac{1}{3}$

The area is half their product:

$\displaystyle A = \frac{1}{2} bh = \frac{1}{2} \cdot 19 \cdot 6 \frac{1}{3} =60\frac{1}{6}$

It is necessary to find the vertices of the triangle, each of which is a point at which two of the three lines intersect.

The two axes intersect at the origin, making this one vertex.

The other two points of intersection are the intercepts of the line of equation $\displaystyle y = 2 x - 15$. Since this equation is in slope-intercept form $\displaystyle y = mx+ b$, where $\displaystyle b$ is the $\displaystyle y$-coordinate of the $\displaystyle y$-intercept, then $\displaystyle b = -15$, and the $\displaystyle y$-intercept is the point $\displaystyle (0, -15)$. The $\displaystyle x$-coordinate of the $\displaystyle x$-intercept, $\displaystyle a$, can be found by setting $\displaystyle y = 0 , x= a$ and solving for $\displaystyle a$:

$\displaystyle 2 x - 15 = y$

$\displaystyle 2 a - 15 = 0$

$\displaystyle 2 a = 15$

$\displaystyle a = \frac{15}{2} = 7 \frac{1}{2}$.

The three vertices are located at $\displaystyle (0,0), \left ( 7\frac{1}{2}, 0 \right ), (0. -15)$

The line in question is shown below, with the bounded triangle shaded in:

The lengths of its legs are equal to the absolute values of the nonzero coordinates of its two intercepts - $\displaystyle a' = 7 \frac{1}{2}, b' = 15$. The area of this right triangle is half their product:

$\displaystyle A = \frac{1}{2} a 'b' = \frac{1}{2} \cdot 7\frac{1}{2} \cdot 15 = 56 \frac{1}{4}$.

The x-intercept is where the line crosses the x-axis. In other words, 

$\displaystyle y=0$

This gives:

$\displaystyle 0=14x+98$

Subtracting 98 from both sides gives:

$\displaystyle -98=14x$

Dividing both sides by 14 gives the final answer:

$\displaystyle x=-7$

It is necessary to find the vertices of the triangle, each of which is a point at which two of the three lines intersect.

The intersection of the $\displaystyle x$-axis - the line $\displaystyle y = 0$ - and the line of the equation $\displaystyle y = x$, is found by noting that if $\displaystyle y = 0$, then, by substitution, $\displaystyle 0 =x$; this point of intersection is at $\displaystyle (0,0)$, the origin. 

The intersection of the $\displaystyle x$-axis and the line of the equation $\displaystyle y = -\frac{4}{3} x + 3$ is found similarly:

$\displaystyle -\frac{4}{3} x + 3 = y$

$\displaystyle -\frac{4}{3} x + 3 = 0$

$\displaystyle -\frac{4}{3} x + 3 - 3 = 0 - 3$

$\displaystyle -\frac{4}{3} x = - 3$

$\displaystyle -\frac{4}{3} x \cdot \left ( -\frac{3} {4} \right ) = - 3 \cdot \left ( -\frac{3} {4} \right )$

$\displaystyle x = \frac{9}{4} = 2 \frac{1}{4}$

This intersection point is at $\displaystyle \left ( 2 \frac{1}{4}, 0 \right )$.

The intersection of the lines with equations $\displaystyle y = x$ and $\displaystyle y = -\frac{4}{3} x + 3$ can be found using the substitution method, setting $\displaystyle y = x$ in the latter equation and solving for $\displaystyle x$:

$\displaystyle x = -\frac{4}{3} x + 3$

$\displaystyle x+ \frac{4}{3} x= -\frac{4}{3} x + 3 + \frac{4}{3} x$

$\displaystyle \frac{7}{3} x=3$

$\displaystyle \frac{3} {7} \cdot \frac{7}{3} x= \frac{3} {7} \cdot 3$

$\displaystyle x= \frac{9} {7} =1\frac{2}{7}$

Since $\displaystyle y = x$, $\displaystyle y =1\frac{2}{7}$, making $\displaystyle \left ( 1\frac{2}{7}, 1\frac{2}{7} \right )$ the point of intersection.

The vertices are at $\displaystyle (0,0),\left ( 2 \frac{1}{4}, 0 \right ),\left ( 1\frac{2}{7}, 1\frac{2}{7} \right )$.

The lines in question are graphed below, and the triangle they bound is shaded:

If we take the horizontal side as the base, its length is seen to be the $\displaystyle x$-coordinate of the $\displaystyle x$-intercept, $\displaystyle 2 \frac{1}{4}$; its (vertical) height is the $\displaystyle y$-coordinate of the opposite vertex, $\displaystyle 1\frac{2}{7}$. The area is half the product of the two, or

$\displaystyle A =\frac{1}{2} \cdot 2 \frac{1}{4} \cdot 1 \frac{2}{7} =1 \frac{25}{56}$

Intercepts occur when a line crosses the $\displaystyle x$-axis or the $\displaystyle y$-axis. When the line crosses the $\displaystyle x$-axis, then $\displaystyle y = 0$ and $\displaystyle x = 3$.  When the line crosses the $\displaystyle y$-axis, then $\displaystyle x = 0$ and $\displaystyle y = 4$. The intercept points are $\displaystyle (0, 4)$ and $\displaystyle (3, 0 )$. So the $\displaystyle y$-intercept is $\displaystyle 4$ and the $\displaystyle x$ intercept is $\displaystyle 3$ and the sum is $\displaystyle 7$.

First, put in slope-intercept form. 

$\displaystyle y = 3x-12-9$

$\displaystyle y=3x-21$. 

To find the $\displaystyle x$-intercept, set $\displaystyle y=0$ and solve for $\displaystyle x$.

$\displaystyle 0=3x-21$

$\displaystyle 3x=21$

$\displaystyle x=7$

To find the x-intercept, set y=0 and solve for x. This gives an answer of x = 7.5.