Solve 0.45/2*(1.2-0.48-y)^2/(1.2-0.48-y)+0.45=3*1/1+frac{y{0.45}}[(x-0.48)y-1/2y^2]

$$\frac { 0.45 } { 2 } \times \frac { ( 1.2 - 0.48 - y ) ^ { 2 } } { ( 1.2 - 0.48 - y ) + 0.45 } = 3 \times \frac { 1 } { 1 + \frac { y } { 0.45 } } [ ( x - 0.48 ) y - \frac { 1 } { 2 } y ^ { 2 } ]$$

Solve for x

$$x=-\frac{2\left(729+10125y-1125y^{2}-6250y^{3}\right)}{375y\left(100y-117\right)}$$

$$y\neq 0\text{ and }y\neq \frac{117}{100}\text{ and }y\neq -\frac{9}{20}$$

Steps for Solving Linear Equation

Expand \frac{0.45}{2} by multiplying both numerator and the denominator by 100.

$$\frac{45}{200}\times \frac{\left(1.2-0.48-y\right)^{2}}{1.2-0.48-y+0.45}=3\times \frac{1}{1+\frac{y}{0.45}}\left(\left(x-0.48\right)y-\frac{1}{2}y^{2}\right)$$

Reduce the fraction \frac{45}{200} to lowest terms by extracting and canceling out 5.

$$\frac{9}{40}\times \frac{\left(1.2-0.48-y\right)^{2}}{1.2-0.48-y+0.45}=3\times \frac{1}{1+\frac{y}{0.45}}\left(\left(x-0.48\right)y-\frac{1}{2}y^{2}\right)$$

Subtract 0.48 from 1.2 to get 0.72.

$$\frac{9}{40}\times \frac{\left(0.72-y\right)^{2}}{1.2-0.48-y+0.45}=3\times \frac{1}{1+\frac{y}{0.45}}\left(\left(x-0.48\right)y-\frac{1}{2}y^{2}\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(0.72-y\right)^{2}.

$$\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.2-0.48-y+0.45}=3\times \frac{1}{1+\frac{y}{0.45}}\left(\left(x-0.48\right)y-\frac{1}{2}y^{2}\right)$$

Subtract 0.48 from 1.2 to get 0.72.

$$\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{0.72-y+0.45}=3\times \frac{1}{1+\frac{y}{0.45}}\left(\left(x-0.48\right)y-\frac{1}{2}y^{2}\right)$$

Add 0.72 and 0.45 to get 1.17.

$$\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.17-y}=3\times \frac{1}{1+\frac{y}{0.45}}\left(\left(x-0.48\right)y-\frac{1}{2}y^{2}\right)$$

Use the distributive property to multiply x-0.48 by y.

$$\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.17-y}=3\times \frac{1}{1+\frac{y}{0.45}}\left(xy-0.48y-\frac{1}{2}y^{2}\right)$$

Use the distributive property to multiply 3\times \frac{1}{1+\frac{y}{0.45}} by xy-0.48y-\frac{1}{2}y^{2}.

$$\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.17-y}=3\times \frac{1}{1+\frac{y}{0.45}}xy-1.44\times \frac{1}{1+\frac{y}{0.45}}y-\frac{3}{2}\times \frac{1}{1+\frac{y}{0.45}}y^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$3\times \frac{1}{1+\frac{y}{0.45}}xy-1.44\times \frac{1}{1+\frac{y}{0.45}}y-\frac{3}{2}\times \frac{1}{1+\frac{y}{0.45}}y^{2}=\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.17-y}$$

Add 1.44\times \frac{1}{1+\frac{y}{0.45}}y to both sides.

$$3\times \frac{1}{1+\frac{y}{0.45}}xy-\frac{3}{2}\times \frac{1}{1+\frac{y}{0.45}}y^{2}=\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.17-y}+1.44\times \frac{1}{1+\frac{y}{0.45}}y$$

Add \frac{3}{2}\times \frac{1}{1+\frac{y}{0.45}}y^{2} to both sides.

$$3\times \frac{1}{1+\frac{y}{0.45}}xy=\frac{9}{40}\times \frac{0.5184-1.44y+y^{2}}{1.17-y}+1.44\times \frac{1}{1+\frac{y}{0.45}}y+\frac{3}{2}\times \frac{1}{1+\frac{y}{0.45}}y^{2}$$

The equation is in standard form.

$$\frac{3y}{\frac{20y}{9}+1}x=\frac{\frac{1458}{125}+162y-18y^{2}-100y^{3}}{-\frac{2000y^{2}}{9}+160y+117}$$

Divide both sides by 3\left(1+\frac{20}{9}y\right)^{-1}y.

$$\frac{\frac{3y}{\frac{20y}{9}+1}x\left(\frac{20y}{9}+1\right)}{3y}=\frac{18\left(729+10125y-1125y^{2}-6250y^{3}\right)}{125\left(1053+1440y-2000y^{2}\right)\times \frac{3y}{\frac{20y}{9}+1}}$$

Dividing by 3\left(1+\frac{20}{9}y\right)^{-1}y undoes the multiplication by 3\left(1+\frac{20}{9}y\right)^{-1}y.

$$x=\frac{18\left(729+10125y-1125y^{2}-6250y^{3}\right)}{125\left(1053+1440y-2000y^{2}\right)\times \frac{3y}{\frac{20y}{9}+1}}$$

Divide \frac{18\left(729+10125y-1125y^{2}-6250y^{3}\right)}{125\left(1053+1440y-2000y^{2}\right)} by 3\left(1+\frac{20}{9}y\right)^{-1}y.

$$x=\frac{2\left(729+10125y-1125y^{2}-6250y^{3}\right)}{375y\left(117-100y\right)}$$