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*= squared (because i cant do the small 2 on the keyboard)

xz*+ xw* =xy* + 5 (solve for x)

2006-09-07 20:08:24 · 12 answers · asked by laurali_love 3 in Science & Mathematics Mathematics

12 answers

xz^2+xw^2=xy^2+5
xz^2+xw^2-xy^2=5
x(z^2+w^2-y^2)=5
x=5/(z^2+w^2-y^2)

Too easy!

2006-09-07 20:14:22 · answer #1 · answered by uselessadvice 4 · 1 0

The key to solving for one variable is isolating it. Recall the addition quality of equality..

if a = b then a - 5 = b - 5

well you have this xz^2 + xw^2 for an 'a' and a xy^2 + 5 for a 'b', right? then it stands to reason that we can subrtract that xy^2 term and it'd end up on the other side.

xz^2 + xw^2 = xy^2 + 5
is now
xz^2 + xw^2 - xy^2 = 5

Now let's use the distributive of numbers to get that x out of each of those terms on the left side.

The distributive property goes like this

a*c + b*c = c*(a+b)

This means that since there's an x in each term, we can just weed it on out of there and put parentheses around the whole group. Sorta like herding in the cattle!

x * (z^2 + w^2 - y^2) = 5

That big boy doesn't look so bad now, does it??

Now all we gotta do is make the x and the 5 switch places.

If we divide by x on both sides, we'll end up with
(z^2 + w^2 - y^2) = 5 / x
and then the inverse is
1 / (z^2 + w^2 - y^2) = x / 5
so multiply by 5 and you get

x = 5 / (z^2 + w^2 - y^2)

2006-09-07 21:10:14 · answer #2 · answered by Neisayer 2 · 0 0

Solving for x

xz² + xw² = xy² + 5

xz² + xw² - xy² = xy² + 5 - xy²

Subtracting - xy² from both sides of the equation

x(z² + w² - y²) = 5

removing the common factor x from the equation

x(z² + w² - y²)/ (z² + w² -y²) = 5/(z² + w² - y²)

Dividing (z² + w² - y²) from both sides of the equation

x = 5/(z² + w² - y²)

2006-09-08 01:45:03 · answer #3 · answered by SAMUEL D 7 · 0 0

xz² + xw² = xy² + 5
xz² + xw² – xy² = 5
x (z² + w² – y²) = 5

x = 5 ÷ (z² + w² – y²)

2006-09-07 20:18:07 · answer #4 · answered by Anonymous · 0 0

LMAO..I looked at the question...then said...5..I may be on the right track.
Purely by accident, unless I am some kind of mathamatical idiot sivante.
That is as opposed to just being a garden variety idiot.
Shine on

2006-09-07 20:16:38 · answer #5 · answered by Comfortably Numb 3 · 0 0

if you don't understand Dan's answer, ^2 means squared. When typing on the computer use the carrot (is that right?) sign ^ followed by the power.

2006-09-07 20:14:55 · answer #6 · answered by Anonymous · 1 0

(x*y)+z*=w*+5
x*=y+z+w+5
x=4

this is an advanced finite mathematics formula which significantly simplifes the problem ..


now try to use this as a hlepfull way to solve your problem effectively

2006-09-07 20:15:52 · answer #7 · answered by Anonymous · 0 1

I hope you at least worked a little on this problem yourself before asking everyone else to do your homework for you...

2006-09-07 20:15:47 · answer #8 · answered by Anonymous · 0 0

x(z^2+w^2-y^2)=5
x=5/(z^2+w^2-y^2)

2006-09-07 20:11:54 · answer #9 · answered by dan 4 · 1 0

x(z*+w*-y*)=5
x=5/(z*+w*-y*)

2006-09-07 20:17:04 · answer #10 · answered by fozerol 1 · 1 0

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