Answer:
x=10
Step-by-step explanation:
8x+10=6x+30
8x-6x=30-10
2x=20
x=10
so first off, let's simplify both equations, starting off by multiplying both sides by the LCD of all fractions, to do away with the denominators.
![\bf \cfrac{10(x-y)-4(1-x)}{3}=y\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{10(x-y)-4(1-x)=3y} \\\\\\ 10x-10y-4+4x=3y\implies \boxed{14x-13y=4} \\\\[-0.35em] ~\dotfill\\\\ 7+x-\cfrac{x-3y}{4}=2x-\cfrac{y+5}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( 7+x-\cfrac{x-3y}{4} \right)=12\left( 2x-\cfrac{y+5}{3} \right)} \\\\\\ 84+12x-3(x-3y)=24x-4(y+5) \\\\\\ 84+12x-3x+9y=24x-4y-20\implies \boxed{-15x+13y=-124}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B10%28x-y%29-4%281-x%29%7D%7B3%7D%3Dy%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B3%7D%7D%7B10%28x-y%29-4%281-x%29%3D3y%7D%20%5C%5C%5C%5C%5C%5C%2010x-10y-4%2B4x%3D3y%5Cimplies%20%5Cboxed%7B14x-13y%3D4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207%2Bx-%5Ccfrac%7Bx-3y%7D%7B4%7D%3D2x-%5Ccfrac%7By%2B5%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B12%7D%7D%7B12%5Cleft%28%207%2Bx-%5Ccfrac%7Bx-3y%7D%7B4%7D%20%5Cright%29%3D12%5Cleft%28%202x-%5Ccfrac%7By%2B5%7D%7B3%7D%20%5Cright%29%7D%20%5C%5C%5C%5C%5C%5C%2084%2B12x-3%28x-3y%29%3D24x-4%28y%2B5%29%20%5C%5C%5C%5C%5C%5C%2084%2B12x-3x%2B9y%3D24x-4y-20%5Cimplies%20%5Cboxed%7B-15x%2B13y%3D-124%7D)
now, let's do some elimination on those two simplified equations.
![\bf \begin{array}{cllcl} 14x&-13y&=&4\\ -15x&+13y&=&-124\\\cline{1-4} -x&&=&-120 \end{array}~\hfill x=\cfrac{-120}{-1}\implies \blacktriangleright x=120 \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{14(120)-13y=4}\implies 1680-13y=4\implies 1680-4=13y \\\\\\ 1676=13y\implies \blacktriangleright \cfrac{1676}{13}=y \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left( 120~,~\frac{1676}{13} \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bcllcl%7D%2014x%26-13y%26%3D%264%5C%5C%20-15x%26%2B13y%26%3D%26-124%5C%5C%5Ccline%7B1-4%7D%20-x%26%26%3D%26-120%20%5Cend%7Barray%7D~%5Chfill%20x%3D%5Ccfrac%7B-120%7D%7B-1%7D%5Cimplies%20%5Cblacktriangleright%20x%3D120%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%201st%20equation%7D%7D%7B14%28120%29-13y%3D4%7D%5Cimplies%201680-13y%3D4%5Cimplies%201680-4%3D13y%20%5C%5C%5C%5C%5C%5C%201676%3D13y%5Cimplies%20%5Cblacktriangleright%20%5Ccfrac%7B1676%7D%7B13%7D%3Dy%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cleft%28%20120~%2C~%5Cfrac%7B1676%7D%7B13%7D%20%5Cright%29~%5Chfill)
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Answer:
B..-10
Step-by-step explanation:
this isn't wrong I promise!
You have 3 spheres inside a cylinder like for a can of tennis balls. Notice the radius of the cylinder is 3 cm, the same as the radius of the spheres, so the spheres just fit width wise, and 3 spheres height would be 6 radius lengths or 18, which is the same as the cylinder height, so the three spheres are packed to the best fit.
The volume not occupied by the spheres will be the total cylinder volume take away the three spheres volumes.
The cyinder volume with height h=18 and radius r=3 will be approx..

The volume of 3 spheres with radius r=3 will be

The volume unoccupied by the spheres inside the cylinder is the difference in those two volumes..
508.68 - 339.12 = 169.56 or 170 cubic centimeters