Answer:
whats the question
Step-by-step explanation:
Answer:
y = 84
Step-by-step explanation:
1) add 3 to both sides of your equation to cancel out the -3
end up with: (1/3)y + 14 = (1/2)y
2) multiply both sides of your equation by 2 to cancel out the (1/2)
end up with: (2/3)y + 28 = y
3) subtract (2/3)y from both sides of the equation to cancel out the positive (2/3)y
end up with: 28 = (1/3)y
4) multiply both sides of the equation by 3 to cancel out the (1/3)
end up with: 84 = y
Im guessing your saying how do you write 1.92 x 10 when you multiply you dont need to line up your decimals but it would lookm something like this:
1.92
x 10
and if it was 1.92 and 10.0 it would look like this
1.92
x 10.0
the deimals dont line up because you are multipling but if you were adding or subtrcating then you would have to line it up but not in this case
Make a substitution:
Then the system becomes
![\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu-v%7D%2B%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu-v%7D-%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
Simplifying the equations gives
![\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
which is to say,
![\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%5Ctimes4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D)
![\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%5B3%5D%7B%5Cleft%28%5Cdfrac%20uv%5Cright%29%5E4%7D%3D81)
Substituting this into the new system gives
![\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7B%28%5Cpm27v%29%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cimplies%5Cdfrac1%7Bv%5E2%7D%3D1%5Cimplies%20v%3D%5Cpm1)
Then
(meaning two solutions are (7, 13) and (-7, -13))
Answer:
Step-by-step explanation:
