Solve the following system using elimination:
{7 x + 2 y = -19 | (equation 1)
{2 y - x = 21 | (equation 2)
Add 1/7 × (equation 1) to equation 2:
{7 x + 2 y = -19 | (equation 1)
{0 x+(16 y)/7 = 128/7 | (equation 2)
Multiply equation 2 by 7/16:
{7 x + 2 y = -19 | (equation 1)
{0 x+y = 8 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{7 x+0 y = -35 | (equation 1)
{0 x+y = 8 | (equation 2)
Divide equation 1 by 7:
{x+0 y = -5 | (equation 1)
{0 x+y = 8 | (equation 2)
Collect results:
Answer: {x = -5, y = 8
The mode is whichever one has the most so the mode would be 7. And the range is the highest number to the lowest. So 9-4 which is 5
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Answer:
5x -y = -37
Step-by-step explanation:
One way to find the coefficients A and B is to use the differences of the x- and y-coordinates:
A = Δy = y2 -y1 = 2 -(-3) = 5
B = -Δx = -(x2 -x1) = -(-7 -(-8)) = -1
Then the constant C can be found using either point.
5x -y = 5(-7) -2 = -37
The equation of the line is ...
5x -y = -37
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<em>Additional comment</em>
This approach comes from the fact that the slope of a line is the same everywhere.

The "standard form" requires that A be positive, so we chose point 1 and point 2 to make sure that was the case.
Find the perpendicular line then find the intersection then find the point
perpendicular lines have slopes that are perpendicular
the slopes multiply bo -1
y=mx+b
m=slope
y=2x-3
2 is slope
2 times what=-1
what=-1/2
the equation is
y-3=-1/2(x-8) or
y=(-1/2)x+7
find intersection
at (4,5)
distance bwetweeen (8,3) and (4,5)
D=
D=
D=
D=

D=2√5
distance= 2√5