Answer: The solution for the system is (2, -7)
Step-by-step explanation:
Ok, here we have linear relationships.
A linear relationship can be written as:
y = a*x + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
In this case, we have two lines:
ya, that passes through:
(-8, -5) and (-3, -6)
Then the slope is:
a = (-6 - (-5))/(-3 - (-8)) = (-6 + 5)/(-3 + 8) = -1/5
now, knowing one of the points like (-3, - 6) we can find the value of b.
y(x) = (-1/5)*x + b
y(-3) = -6 = (-1/5)*-3 + b
-6 = 3/5 + b
b = -6 - 3/5 = -33/5
then the first line is:
ya = (-1/5)*x -33/5
For the second line, we know that it passes through the points:
(-8, -15) and (-3, -11)
Then the slope is:
a = (-11 - (-15))/(-3 -(-8)) = (-11 + 15)/(-3 + 8) = 4/5
The our line is:
y(x) = (4/5)*x + b
and for b, we do the same as above, using one of the points, for example (-3, -11)
y(-3) = -11 = (4/5)*-3 + b
b = -11 + 12/5 = -(55 + 12)/5 = -43/5
then:
yb = (4/5)*x - 43/5.
Ok, our system of equations is:
ya = (-1/5)*x -33/5
yb = (4/5)*x - 43/5.
To solve this, we suppose ya = yb
then:
(-1/5)*x + -33/5 = (4/5)*x - 43/5.
-33/5 + 43/5 = (4/5)*x + (1/5)*x
10/5 = 2 = (4/5 + 1/5)*x = x
2 = x
now we evaluate x = 2 in one of the lines:
ya = (-1/5)*2 -33/5 = -2/5 - 33/5 = -35/5 = -7
Then the lines intersect at the point (2, - 7), which is the solution for the system.
