\(\frac{x\left(y^2-3y^2\right)}{2x}=\frac{y^3}{-\left(4x+y\right)}\)
for both x and y...
+36919 I'll do the first one...solve for 'x'
First simplify the expression
x(-2y^2) / 2x = y^3/(-(4x+y) simplify the left side (the 'x' cancels)
-2y^2/2 =Y^3/(-(4x+y) Now the 2's cancel on the left
-y^2 = y^3/ (-(4x+y)) Divide both sides by y^2
-1 = y/(-(4x+y)) Multiply both sides by -(4x +y)
4x+y = y Subtract y from both sides
4x=0 Divide both sides by 4
x=0
+1904 Solve for x.
\(\frac{x({y}^{2}-3{y}^{2})}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(\frac{x(-2{y}^{2})}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(\frac{-2x{y}^{2}}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-\frac{2x{y}^{2}}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-\frac{1x{y}^{2}}{x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-\frac{x{y}^{2}}{x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-1{y}^{2}=\frac{{y}^{3}}{-(4x+y)}\)
\(-{y}^{2}=\frac{{y}^{3}}{-(4x+y)}\)
\(-{y}^{2}=-\frac{{y}^{3}}{(4x+y)}\)
\(-{y}^{2}\times(4x+y)=-\frac{{y}^{3}}{(4x+y)}\times(4x+y)\)
\(-{y}^{2}\times(4x+y)=-{y}^{3}\)
\(\frac{-{y}^{2}\times(4x+y)}{-{y}^{2}}=\frac{-{y}^{3}}{-{y}^{2}}\)
\(\frac{-{y}^{2}\times(4x+y)}{-{y}^{2}}=\frac{{y}^{3}}{{y}^{2}}\)
\(\frac{-{y}^{2}\times(4x+y)}{-{y}^{2}}={y}^{3-2}\)
\(\frac{-{y}^{2}\times(4x+y)}{-{y}^{2}}={y}^{1}\)
\(\frac{-{y}^{2}\times(4x+y)}{-{y}^{2}}=y\)
\(4x+y=y\)
\(4x+y-y=y-y\)
\(4x+0=y-y\)
\(4x=y-y\)
\(4x=0\)
\(\frac{4x}{4}=\frac{0}{4}\)
\(1x=\frac{0}{4}\)
\(x=\frac{0}{4}\)
\(x=0\)
Check to see if \(x=0\) is the correct solution.
\(\frac{x({y}^{2}-3{y}^{2})}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(\frac{0({y}^{2}-3{y}^{2})}{2\times0}=\frac{{y}^{3}}{-(4\times0+y)}\)
\(\frac{0(-{2y}^{2})}{2\times0}=\frac{{y}^{3}}{-(4\times0+y)}\)
\(\frac{0}{2\times0}=\frac{{y}^{3}}{-(4\times0+y)}\)
\(\frac{0}{0}=\frac{{y}^{3}}{-(4\times0+y)}\)
\(\frac{0}{0}=\frac{{y}^{3}}{-(0+y)}\)
\(\frac{0}{0}=\frac{{y}^{3}}{0-y}\)
\(\frac{0}{0}=\frac{{y}^{3}}{-y}\)
\(\frac{0}{0}=-\frac{{y}^{3}}{y}\)
\(\frac{0}{0}=-{y}^{3-1}\)
\(\frac{0}{0}=-{y}^{2}\)
\(Undefined=-{y}^{2}\)
Because both sides are not equal to each other \(x=0\) is not a solution which means that there is no solution to this problem.
Solve for y.
\(\frac{x({y}^{2}-3{y}^{2})}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(\frac{x(-2{y}^{2})}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(\frac{-2x{y}^{2}}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-\frac{2x{y}^{2}}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-\frac{1x{y}^{2}}{x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-\frac{x{y}^{2}}{x}=\frac{{y}^{3}}{-(4x+y)}\)
\(-1{y}^{2}=\frac{{y}^{3}}{-(4x+y)}\)
\(-{y}^{2}=\frac{{y}^{3}}{-(4x+y)}\)
\(-{y}^{2}=-\frac{{y}^{3}}{(4x+y)}\)
\(-{y}^{2}\times(4x+y)=-\frac{{y}^{3}}{(4x+y)}\times(4x+y)\)
\(-{y}^{2}\times(4x+y)=-{y}^{3}\)
\(-4{y}^{2}x-{y}^{3}=-{y}^{3}\)
\(-4{y}^{2}x-{y}^{3}+{y}^{3}=-{y}^{3}+{y}^{3}\)
\(-4{y}^{2}x-0=-{y}^{3}+{y}^{3}\)
\(-4{y}^{2}x=-{y}^{3}+{y}^{3}\)
\(-4{y}^{2}x=0\)
\(\frac{-4{y}^{2}x}{-4}=\frac{0}{-4}\)
\(1{y}^{2}x=\frac{0}{-4}\)
\({y}^{2}x=\frac{0}{-4}\)
\({y}^{2}x=-\frac{0}{4}\)
\({y}^{2}x=0\)
\(\frac{{y}^{2}x}{x}=\frac{0}{x}\)
\(1{y}^{2}=\frac{0}{x}\)
\({y}^{2}=\frac{0}{x}\)
\({y}^{2}=0\)
\(\sqrt{{y}^{2}}=\sqrt{0}\)
\(y=\sqrt{0}\)
\(y=0\)
Check to see if \(y=0\) is the correct solution.
\(\frac{x({y}^{2}-3{y}^{2})}{2x}=\frac{{y}^{3}}{-(4x+y)}\)
\(\frac{x({0}^{2}-3\times{0}^{2})}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(\frac{x(0-3\times{0}^{2})}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(\frac{x(0-3\times0)}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(\frac{x(0-0)}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(\frac{x(0)}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(\frac{0x}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(\frac{0}{2x}=\frac{{0}^{3}}{-(4x+0)}\)
\(0=\frac{{0}^{3}}{-(4x+0)}\)
\(0=\frac{0}{-(4x+0)}\)
\(0=\frac{0}{-4x-0}\)
\(0=\frac{0}{-4x}\)
\(0=-\frac{0}{4x}\)
\(0=0\)
Since both sides are equal \(y=0\) is a correct solution to this problem.
+36919 Nice job GJ....I just solved the equation for 'x' as requested... Obviously x=0 will be 'undefined' with a '0' in the denominator........Which only says the equation is INVALID for x=0 (but that is not what they asked).......I did solve for y too, but did not post it, YOU did !
Strong work!
