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# system of two equations

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I have two equations: 4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2 Can you solve this for x and y, and if yes, how?

Guest Apr 4, 2015

#4
+84159
+10

4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2

Rearanging both equations, we have

x = 8/(5- y)^2      and  x = 2/(3-y)^2

This implies that

8/(5- y)^2  =  2/(3-y)^2      divide both sides by 2

4/(5 - y)^2  = 1/(3-y)^2       cross-multiply  and simplify

4(y^2 - 6y + 9) = y^2 - 10y + 25

4y^2 - 24y + 36  = y^2 - 10y + 25

3y^2 - 14y + 11  = 0    factor

(3y -11) (y - 1) = 0     set each factor to 0

y = 11/3    and y  = 1

And when y = 11/3, x =  9/2

And when y = 1, x = 1/2

Here's the graph of both    x = 8/(5- y)^2      and  x = 2/(3-y)^2   and the points of intersection

https://www.desmos.com/calculator/ky62ryvrhs

P.S.- The graphs look as though they might intersect below the y axis at some point.....in fact..they don't !!!

CPhill  Apr 4, 2015
Sort:

#1
+17711
+10

You can start by doing this:

4 = 0.5x(5 - y)²

--->   8  =  x(5 - y)²   --->   8/x  = (5 - y)²   --->   ± √(8)/√(x)  =  5 - y

--->   y  =  5 ± √(8)/√(x)   --->   y  =  5 ± 2√(2)/√(x)

similarly:

1  =  0.5x(3 - y)²

--->   2  =  x(3 - y)²   --->   2/x  = (3 - y)²   --->   ± √(2)/√(x)  =  3 - y

--->   y  =  3 ± √(2)/√(x)

Choosing one from each group and setting them equal to each other:

5 - 2√(2)/√(x)  =  3 - √(2)/√(x)

--->   2  = 2√(2)/√(x) - √(2)/√(x)   --->   2  =  √(2)/√(x)   --->   2√(x)  =  √(2)   --->   √(x)  =  √(2)/2

--->   x  =  2/4   --->   x  =  1/2

When  x  =  1/2, y  =  3 -  √(2)/√(1/2)   --->   y  =  3 - √(4)   --->   y  =  3 - 2  =  1

So, one answer is (0.5, 1).

But, I choose only one possibility from each. Now, it's up to you to choose the other 3 possibilities (not all may work).

This is rather sketchy; if you want further help, please ask.

geno3141  Apr 4, 2015
#2
+5

Thank you very much! I tried to isolate x, but that wasn't very helpful, ended up with a long division.
You're my hero for today! :)

Guest Apr 4, 2015
#3
+17711
+5

I'm very glad to be able to help ...

geno3141  Apr 4, 2015
#4
+84159
+10

4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2

Rearanging both equations, we have

x = 8/(5- y)^2      and  x = 2/(3-y)^2

This implies that

8/(5- y)^2  =  2/(3-y)^2      divide both sides by 2

4/(5 - y)^2  = 1/(3-y)^2       cross-multiply  and simplify

4(y^2 - 6y + 9) = y^2 - 10y + 25

4y^2 - 24y + 36  = y^2 - 10y + 25

3y^2 - 14y + 11  = 0    factor

(3y -11) (y - 1) = 0     set each factor to 0

y = 11/3    and y  = 1

And when y = 11/3, x =  9/2

And when y = 1, x = 1/2

Here's the graph of both    x = 8/(5- y)^2      and  x = 2/(3-y)^2   and the points of intersection

https://www.desmos.com/calculator/ky62ryvrhs

P.S.- The graphs look as though they might intersect below the y axis at some point.....in fact..they don't !!!

CPhill  Apr 4, 2015

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