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

This gives:

Subtracting 98 from both sides gives:

Dividing both sides by 14 gives the final answer:

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 -axis - the line  - and the line of the equation , is found by noting that if , then, by substitution, ; this point of intersection is at , the origin. 

The intersection of the -axis and the line of the equation  is found similarly:

This intersection point is at .

The intersection of the lines with equations  and  can be found using the substitution method, setting  in the latter equation and solving for :

Since , , making  the point of intersection.

The vertices are at .

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 -coordinate of the -intercept, ; its (vertical) height is the -coordinate of the opposite vertex, . The area is half the product of the two, or

Intercepts occur when a line crosses the -axis or the -axis. When the line crosses the -axis, then  and .  When the line crosses the -axis, then  and . The intercept points are  and . So the -intercept is  and the  intercept is  and the sum is .

First, put in slope-intercept form. 

. 

To find the -intercept, set  and solve for .

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

The formula for the equation of a circle is:

(x-h) ² + (y-k)² = r²

Where (h,k) is the center of the circle.

h = 0 and k = 4

and diameter = 6 therefore radius = 3

(x-0) ² + (y-4)² = 3²

x² + (y-4)² = 9

The general equation of a circle is (x – h)² + (y – k)² = r², where (h, k) represents the location of the circle's center, and r represents the length of its radius. 

Circle A first has the equation of (x – 4)² + (y + 3)² = 29. This means that its center must be located at (4, –3), and its radius is √29.

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is 2√29.

Now, that we have circle A's new center and radius, we can write its general equation using (x – h)² + (y – k)² = r².

(x – (–2))² + (y – 2)² = (2√29)² = 2²(√29)² = 4(29) = 116.

(x + 2)² + (y – 2)² = 116.

The answer is (x + 2)² + (y – 2)² = 116.

We are trying to find an equation for all of the points that are the same distance (5 units) from (–3, 6). The locus of all points equidistant from a single point is a circle. In other words, we need to find an equation of a circle. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5. 

The standard form of a circle is given below:

(x – h)² + (y – k)² = r², where the center is located at (h, k) and r is the length of the radius.

In this case, h will be –3, k will be 6, and r will be 5.

(x – (–3))² + (y – 6)² = 5²

(x + 3)² + (y – 6)² = 25

The answer is (x + 3)² + (y – 6)² = 25.

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

4x + 3 = 5x + 44; 3 = x + 44; –41 = x

To find the y-coordinate, substitute into one of the equations.  Let's use y₁:

y = 4 * –41 + 3 = –164 + 3 = –161

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at (x₀, y₀) is:

(x - x₀)² + (y - y₀)² = r²

For our data, this means that our equation is:

(x + 41)² + (y + 161)² = 12² or (x + 41)² + (y + 161)² = 144