If 4x-7y=3 and xy=5, find the value of 16x²+49y²
Plz Reply me fast
Answers
Answered by
4
4x-7y=3
Squaring both sides
16x^2+49y^2-56xy=9
16x^2+49y^2-56×5=9
16x^2+49y^2-280=9
16x^2+49y^2=289
Squaring both sides
16x^2+49y^2-56xy=9
16x^2+49y^2-56×5=9
16x^2+49y^2-280=9
16x^2+49y^2=289
Anonymous:
can you tell me that from where 56xy has came
чup
ѕhє díd (4х -3ч)² ѕσ ѕhє gσt..16х²+49ч² -2.4х.7ч
αnd thuѕ ít cσmєѕ -56хч
Answered by
4
hope u like my process...
____________________
4х -7ч = 3. __хч = 5
fσrmulα uѕєd : α² + в² = (α-в)²+2αв
----------------------------------------------------
(4х)²+(7ч) = (4х-7ч)² +2×4х×7ч
σr, 16х² +49ч² =3²+2×4×7×5
ѕσ, 16х² +49ч²= 9+280=289 __αnѕwєr
hσpє thíѕ íѕ ur rєquírєd αnѕwєr.
prσud tσ hєlp u.
____________________
4х -7ч = 3. __хч = 5
fσrmulα uѕєd : α² + в² = (α-в)²+2αв
----------------------------------------------------
(4х)²+(7ч) = (4х-7ч)² +2×4х×7ч
σr, 16х² +49ч² =3²+2×4×7×5
ѕσ, 16х² +49ч²= 9+280=289 __αnѕwєr
hσpє thíѕ íѕ ur rєquírєd αnѕwєr.
prσud tσ hєlp u.
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