I'm assuming that you are solving for the variables, x and y.
Given:
(a) x - y = 4
(b) x^2 + 2xy + y^2 = 64
Recognize that (b) can be factored:
x^2 + 2xy + y^2 = 64
=> (x + y)^2 = 64
=> x + y = +/-8...(c)
Rearranging (a) will give you: x = y + 4...(d)
Now substitute (d) into (c) to solve for y:
x + y = +/-8
=> (y + 4) + y = +/-8
=> 2y + 4 = 8 or -8
=> 2y = 4 or -12
=> y = 2 or -6
Substitute the y-values back into (a) to solve for x:
x - y = 4
=> x - (2) = 4 or x - (-6) = 4
=> x -2 = 4 or x + 6 = 4
=> x = 6 or -2
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So this system of equations actually has two answers:
When x = 6, y = 2, and
When x = -2, y = -6
The solution can also be expressed like this: (6, 2), (-2, -6)
2007-12-07 05:13:27
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answer #1
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answered by Aquaboy 6
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1) x-y = 4
2) x^2 + 2xy + y^2 = 64
Factor #2:
(x+y)(x+y) = 64
(x+y)^2 = 64
(x+y) = +/-8
So now we have 3 possible equations:
1) x-y = 4
2) x+y = 8
3) x+y = -8
We can solve #1 for x and substitute:
x - y = 4
x = 4+y
2) x + y = 8; (4+y) + y = 8; 4 + 2y = 8; 2y = 4; y = 2
3) x + y = -8; 4 + 2y = -8; 2y = -12; y = -6
This gives us two possible solutions for y: 2 and -6
Substitute these into any original equation and solve for x:
x - y = 4
x - 2 = 4
x = 6
x - (-6) = 4
x + 6 = 4
x = -2
So our two solutions are:
x = 6, y = 2
x = -2, y = -6
or
(6,2) and (-2, -6)
Hope this helps! :)
2007-12-07 05:17:06
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answer #2
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answered by disposable_hero_too 6
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From x-y=4,
we could find x=4+y.
From x^2+2xy +y^2=64,
we could solve it like this:
(x+y)^2=64
x+y=√64
x+y=8
Subs x=4+y into x+y=8,
(4+y)+y=8
4+2y=8
2y=4
y=2
∴x=6,y=2
2007-12-07 05:13:11
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answer #3
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answered by Simple and Complicated 1
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if youre fixing for x, its straight forward, first upload y to the two components which provide you with x+2xy=4+y then element out the x on the left component of the equation, you have got x(a million+2y) = 4+y they devide the two components by ability of a million+2y the respond is (4+y)/(a million+2y) that's an identical element as (4+y)/(2y+a million)
2016-11-13 23:42:25
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answer #4
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answered by ? 4
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first term tells you x = y+4 or y = x-4
substitute that in second term you get
x^2+2x(x-4)+(x-4)^2=64
expand that to
x^2 + 2x^2 - 8x + x^2 - 8x +16 = 64
sort it out to make
4x^2 - 16x = 48
now estimate and calculate
if x =2 then 4x^2 - 16x = 16 - 32 = -16
if x = 4 then 4x^2 - 16x = 64 - 64 = 0
if x = 6 then 4x^2 - 16x = 144 - 96 = 48 so x = 6 and y = 2
2007-12-07 05:20:36
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answer #5
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answered by derbydolphin 7
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x - y = 4;
x = y + 4, &
y = x - 4,
x = 6, & y = 2. (ans).
check: [6 -2 = 4].
x^2 + 2xy + y^2 = 64;
x = 4, & y = 4. (ans).
check: [4*4] + [2*4*4] + [4*4] = 64.
(I couldn't factorise).
2007-12-07 05:22:17
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answer #6
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answered by BB 7
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One obvious way to solve this is to solve for X in the first equation and then use substitution (wherever you see X in the second equation, insert its equivalent Y expression).
2007-12-07 05:15:10
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answer #7
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answered by Anonymous
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x-y =4
56-52=4
or
128-124=4
or
1687-1683=4
or
19846-19842=4
or
567894-567890=4
It could be any of these, or many others.
If you haven't been given a value for either x or y then the correct answer would be x-y=4
2007-12-07 05:17:06
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answer #8
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answered by Blonde hootie 3
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im like blokheed ur not alone
2007-12-08 08:36:32
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answer #9
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answered by Laura 6
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You're not alone - neither can I.
And neither can they ↓↓↓
Ha ha ha.
2007-12-07 05:02:42
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answer #10
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answered by Blokheed 5
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