Happy New year family

Please can someone help with solutions to these

X²+y²=34
xy=15 find the value of x in the simultaneous equation

X = 3
Y = 5
Or
X = 5
Y = 3. More instances will be needed to ascertain the real value of x but either ways works fine ATM.

Ok

X= 5 or 3

Maks27

X² + Y² = 34
Y =√(34 - X²) since YX=15
Then √(34 - X²) * X = 15
√(34 - X²) = 15/X
Taking the square of both sides
34 - X² = 225/X²
Multiplying throughout by X²
34X² - X⁴ = 225
Rearranging
X⁴ - 34X² + 225 = 0
Solving for X² using the quadratic formula
X² = (-b ± √(b² - 4ac))/2a
Where a=1, b=-34 and c=225
X² = {-(-34) ± √[(-34)² - 4(1)(225)]}/2(1)
X² = 17 ± √64
X² = 17 ± 8
Hence
X² = 9 or 25
Therefore
X = 3 or 5 (by taking root of both sides)

X =3
Y=5 ans
Explanation:
X²+y²=34
XY=15
Therefore 3²+5²=34
9+25=34
√9=3.
X=3 ans

Maks27:
Happy New year family

Please can someone help with solutions to these

X²+y²=34
xy=15 find the value of x in the simultaneous equation

DOWNLOAD the app PHOTOMATH and solve this in 30 secs

Maks27:
Happy New year family

Please can someone help with solutions to these

X²+y²=34
xy=15 find the value of x in the simultaneous equation

either x = 5 and y = 3

or x =3 and y = 5.