A car depreciated $1000 the first year it was owned and driven. In years 2 and
2 answers:
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Answer:
Ok, we have a system of equations:
6*x + 3*y = 6*x*y
2*x + 4*y = 5*x*y
First, we want to isolate one of the variables,
As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:
(6*x + 3*y)/(2*x + 4*y) = 6/5
now we isolate one off the variables:
6*x + 3*y = (6/5)*(2*x + 4*y) = (12/5)*x + (24/5)*y
x*(6 - 12/5) = y*(24/5 - 3)
x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y
Now we can replace it in the first equation:
6*x + 3*y = 6*x*y
6*(0.5*y) + 3*y = 6*(0.5*y)*y
3*y + 3*y = 3*y^2
3*y^2 - 6*y = 0
Now we can find the solutions of that quadratic equation as:
So we have two solutions
y = 0
y = 2.
Suppose that we select the solution y = 0
Then, using one of the equations we can find the value of x:
2*x + 4*0 = 5*x*0
2*x = 0
x = 0
(0, 0) is a solution
if we select the other solution, y = 2.
2*x + 4*2 = 5*x*2
2*x + 8 = 10*x
8 = (10 - 2)*x = 8x
x = 1.
(1, 2) is other solution
