find the value of x cube and y cube and xy when x+y=5 and x-y=3
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Answered by
3
Here we can make it by elimination method as follows
x+y=5 - eq1
x-y=3 - eq2
so, by equating we have 2x =8
Thus, x=4 .Then its cube 64
Now as we found x so by putting its value in any equation we wil have y = 1 so its cube =1
x+y=5 - eq1
x-y=3 - eq2
so, by equating we have 2x =8
Thus, x=4 .Then its cube 64
Now as we found x so by putting its value in any equation we wil have y = 1 so its cube =1
SamRaiden:
just a sec
Done
first re arrange it
ie (x+4)(x-5)(x+6)(x-7=504) After that it will be like
{x^2-x-20}{x^2-x-42}
let (x^2 - x )=a then it will be like _(a-20)(a-42)=504
now bro try it your self
after it u will get (a) ie (x^2 - x ) evaluate it u wil get ir answer
it will be there will be 2 case and root will be -2,3-7,8
bbye
Answered by
1
x+y=5---------1
x-y=3-----------2
---------
2x=8
x=4
on putting the value of x in eq. 2
4-y=3
-y=3-4
-y=-1
y=1
x=4
xcube=4cube (on cubing both sides)
xcube=64
y=1
ycube=1 cube (on cubing both sides)
ycube=1
x-y=3-----------2
---------
2x=8
x=4
on putting the value of x in eq. 2
4-y=3
-y=3-4
-y=-1
y=1
x=4
xcube=4cube (on cubing both sides)
xcube=64
y=1
ycube=1 cube (on cubing both sides)
ycube=1
Thank you
welcome
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