x^2+y^2+10=2√2x+4√2y then find value of (x+y)
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Answer:
Step-by-step
{2^{x}+2^{y}=10, x+y=4┊⇒{2^{x}+2^{4-x}=10, y=4-x┊
^{x}+16=0
t=2^{x}>0⇒t²-10t+16=0
Δ=100-64=36
t_{1,2}=((10±6)/2)⇒t₁=8=2³,t₂=2=2¹
⇒x₁=3,y₁=1
x₂=1,y₂=3
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Timothy Christian Liu, Mathematics was my forte back in high school.
Answered 5 years ago · Author has 2.4K answers and 9.3M answer views
Originally Answered: How can one solve the simultaneous equations 2^x +2^y =10 and x+y=4 ?
Substitute y = 4 - x into the first equation,
So 2^x + 2^(4-x) = 10. This can also be rewritten as 2^x + 2^4 / 2^x = 10
Let a = 2^x, so we can simplify this equation to a^2 - 10a + 16 = 0.
Solve this quadratic equation to obtain a = 2 or 8, hence x = 1 or 3, while y = 3 (when x = 1) or 1 (when x = 3) respectively.
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Rohan Talwar, studied at Bal Bharati Public School
Answered 4 years ago
from observation to get the sum 10 from exp of 2 the possible answer are (8,2) or (2,8) ,So we can say that the possible (x,y) =(3,1) or (1,3)
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Easy answers are staring you in the face and it is more interesting to solve the problem systematically. When an answer has been found, the x and y values can be interchanged and we can therefore get everything we want by assuming that x>y. Square both sides of the first equation and subtract 4 times 2^(x+y). This gives (2^x)^2 - 2*(2^x)*(2^y) + (2^y)^2 = 36. This gives 2^x-2^y =6. Solve this together with the first given equation and we find 2^x=8 and 2^y=2 or x=3, y=1.
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y = 4 - x. Thus
2^x + 2^(4 - x) = 10
2^x + 2^4/2^x = 10
Let z = 2^x. Then
z + 16/z = 10
z^2 + 16 = 10z
z^2 - 10z + 16 = 0
(z - 2)(z - 8) = 0
z = 2 or 8
2^x = 2 or 8
So x = 1 or 3
y = 4 - 1 = 3 or 4 - 3 = 1
(x, y) = (1, 3), (3, 1)
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L Viswanathan, Loves Math & Physics, enjoys Sanskrit
Answered March 1, 2021 · Author has 2.5K answers and 513.6K answer views
x + y = 4. ==> y = 4-x
2^x + 16/(2^x) = 10.
Let 2^x = u
u + (16/u) - 10 = 0
==> u^2 - 10u + 16 = (u - 8)(u - 2) = 0
==> u = 2^3 and 2^1
(x, y) = (3, 1) or (1, 3)
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Jonathan E. Segal
Answered March 10, 2021 · Author has 185 answers and 14.8K answer views
Here is how I woulD solve this problem:
Take the first equation and multiply both sides by 2^x. When you do, you should have the following:
2^x*(2^x+2^y) = 10*2^x
Therefore, 2^x*2^x + 2^x*2^y = 10*2^x
(2^x)^2+2^(x+y) = 10*2^x
replace x+y with 4
(2^x)^2+2^(4) = 10*2^x
let u = 2^x
u^2 + 16 = 10u
If I solve for u, I get u= 2 or u = 8.
but u = 2^ x
So, 2^x = 2 or 2^ x = 8
The only way that can happen is if x= 0 or x= 3
if x=0, then y = 4.
If x= 3, then y = 1.
The only combination that works for both equations is x = 3 and y = 1.
Answer: {(3, 1)}
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Kurt Mager, Enjoys solving math problems
Answered June 26, 2021 · Author has 3.6K answers and 1.6M answer views
x+y=4⟹y=4−x
2x+2y=10
⟹2x+24−x=10
⟹2x+162x=10
⟹(2x)2−10(2x)+16=0
⟹(2x−2)(2x−8)=0
⟹2x=2,8
x=1 or 3⟹y=3 or 1
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Joseph Yiin
Answered 5 years ago · Author has 110 answers and 102.9K answer views
x + y =4 ==> y = 4-x
2^x + 2^(4-x) = 10
==> 2^x + (2^4)/(2^x) = 10
==> 2^(2*x) + 16 = 10*(2^x)
let z = 2^x
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