DRAW AND LABEL AN EXAMPLE OF TWO PARALLEL LINES THAY ARE PERPENDICULAR TO A THIRD LINE
2 answers:
You might be interested in
Answer:
(x, y) = (2, 5)
Step-by-step explanation:
I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...
3x' -y' = 13/10
x' +2y' = 9/10
Adding twice the first equation to the second, we get ...
2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)
7x' = 35/10 . . . . . . simplify
x' = 5/10 = 1/2 . . . . divide by 7
Using the first equation to find y', we have ...
y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5
So, the solution is ...
x = 1/x' = 1/(1/2) = 2
y = 1/y' = 1/(1/5) = 5
(x, y) = (2, 5)
_____
The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.
<u>At Hayden's:</u>
<u>At Austins:</u>
As we can see in both results, it would take them (4 hours) for the prices to be equaled.
<u>At Austins:</u>
As we can see in both results, it would take them (4 hours) for the prices to be equaled.