Answer:
A linear relation can be written in the slope-intercept form as:
y = a*x + b
Where a is the slope, and b is the y-intercept.
And if a line passes through the points (x₁, y₁) and (x₂, y₂), the slope can be written as:
a = (y₂ - y₁)/(x₂ - x₁)
Now let's look at our lines.
The left one passes through the points (0, -3) and (-2, 0)
Then the slope of this line is:
a = (0 - (-3))/(-2 - 0) = 3/-2 = -(3/2)
Then the line will be something like:
y = -(3/2)*x + b
Knowing that this line passes through the point (0, -3), we know that when x = 0, we must have y = -3
Then:
-3 = (-3/2)*0 + b
-3 = b
So the first linear equation is:
y = -(3/2)*x - 3
Now let's look at the second line, this one passes through the points (0, 2) and (1, 0)
Then the slope of this line is:
a = (0 - 2)/(1 - 0) = -2
And we can write the line as:
y = -2*x + b
Knowing that this line passes through the point (0, 2), we know that when x = 0 we must have y = 0, replacing that we get:
2 = -2*0 + b
2 = b
Then the equation of this line is:
y = -2*x + 2
Now that we know both equations we can write our system as:
y = (-3/2)*x - 3
y = -2*x + 2
To solve it, we need to remember that y = y
then:
(-3/2)*x - 3 = y = -2*x + 2
(-3/2)*x - 3 = -2*x + 2
We can solve this for x.
2*x - (3/2)*x = 2 + 3
(1/2)*x = 5
x = 5((1/2) = 5*2 = 10
And to find the y-value, we need to input this x-value in one of the equations:
y = -2*10 + 2 = -20 + 2 = -18
Then the solution of the system is the point (10, -18)