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2007-12-07 04:58:20 · 10 answers · asked by Anisha 1 in Education & Reference Homework Help

10 answers

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

--------------------------------------------------------------
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 · answer #1 · answered by Aquaboy 6 · 2 1

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 · answer #2 · answered by disposable_hero_too 6 · 2 1

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 · answer #3 · answered by Simple and Complicated 1 · 1 3

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 · answer #4 · answered by ? 4 · 0 0

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 · answer #5 · answered by derbydolphin 7 · 1 3

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 · answer #6 · answered by BB 7 · 1 3

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 · answer #7 · answered by Anonymous · 0 2

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 · answer #8 · answered by Blonde hootie 3 · 0 4

im like blokheed ur not alone

2007-12-08 08:36:32 · answer #9 · answered by Laura 6 · 0 0

You're not alone - neither can I.

And neither can they ↓↓↓

Ha ha ha.

2007-12-07 05:02:42 · answer #10 · answered by Blokheed 5 · 1 4

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