Solve {l}{x(1-2x)-y(1-y)=(y-sqrt{2}x)(y+sqrt{2}x)+3}{2x-(2y-1/4)^2+161/16=(2y+3)(3-2y)}

$$\left\{ \begin{array} { l } { x ( 1 - 2 x ) - y ( 1 - y ) = ( y - \sqrt { 2 } x ) ( y + \sqrt { 2 } x ) + 3 } \\ { 2 x - ( 2 y - \frac { 1 } { 4 } ) ^ { 2 } + 16 \frac { 1 } { 16 } = ( 2 y + 3 ) ( 3 - 2 y ) } \end{array} \right.$$

Solve for x, y

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$y = -\frac{13}{3} = -4\frac{1}{3} \approx -4.333333333$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply x by 1-2x.

$$x-2x^{2}-y\left(1-y\right)=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

Use the distributive property to multiply y by 1-y.

$$x-2x^{2}-\left(y-y^{2}\right)=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

To find the opposite of y-y^{2}, find the opposite of each term.

$$x-2x^{2}-y+y^{2}=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

Consider \left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x-2x^{2}-y+y^{2}=y^{2}-\left(\sqrt{2}x\right)^{2}+3$$

Expand \left(\sqrt{2}x\right)^{2}.

$$x-2x^{2}-y+y^{2}=y^{2}-\left(\sqrt{2}\right)^{2}x^{2}+3$$

The square of \sqrt{2} is 2.

$$x-2x^{2}-y+y^{2}=y^{2}-2x^{2}+3$$

Subtract y^{2} from both sides.

$$x-2x^{2}-y+y^{2}-y^{2}=-2x^{2}+3$$

Combine y^{2} and -y^{2} to get 0.

$$x-2x^{2}-y=-2x^{2}+3$$

Add 2x^{2} to both sides.

$$x-2x^{2}-y+2x^{2}=3$$

Combine -2x^{2} and 2x^{2} to get 0.

$$x-y=3$$

Consider the second equation. Multiply both sides of the equation by 16.

$$16\left(2x-\left(2y-\frac{1}{4}\right)^{2}\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-\frac{1}{4}\right)^{2}.

$$16\left(2x-\left(4y^{2}-y+\frac{1}{16}\right)\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

To find the opposite of 4y^{2}-y+\frac{1}{16}, find the opposite of each term.

$$16\left(2x-4y^{2}+y-\frac{1}{16}\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Use the distributive property to multiply 16 by 2x-4y^{2}+y-\frac{1}{16}.

$$32x-64y^{2}+16y-1+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Multiply 16 and 16 to get 256.

$$32x-64y^{2}+16y-1+256+1=16\left(2y+3\right)\left(3-2y\right)$$

Add -1 and 256 to get 255.

$$32x-64y^{2}+16y+255+1=16\left(2y+3\right)\left(3-2y\right)$$

Add 255 and 1 to get 256.

$$32x-64y^{2}+16y+256=16\left(2y+3\right)\left(3-2y\right)$$

Use the distributive property to multiply 16 by 2y+3.

$$32x-64y^{2}+16y+256=\left(32y+48\right)\left(3-2y\right)$$

Use the distributive property to multiply 32y+48 by 3-2y and combine like terms.

$$32x-64y^{2}+16y+256=-64y^{2}+144$$

Add 64y^{2} to both sides.

$$32x-64y^{2}+16y+256+64y^{2}=144$$

Combine -64y^{2} and 64y^{2} to get 0.

$$32x+16y+256=144$$

Subtract 256 from both sides.

$$32x+16y=144-256$$

Subtract 256 from 144 to get -112.

$$32x+16y=-112$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$x-y=3,32x+16y=-112$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$x-y=3$$

Add y to both sides of the equation.

$$x=y+3$$

Substitute y+3 for x in the other equation, 32x+16y=-112.

$$32\left(y+3\right)+16y=-112$$

Multiply 32 times y+3.

$$32y+96+16y=-112$$

Add 32y to 16y.

$$48y+96=-112$$

Subtract 96 from both sides of the equation.

$$48y=-208$$

Divide both sides by 48.

$$y=-\frac{13}{3}$$

Substitute -\frac{13}{3} for y in x=y+3. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{13}{3}+3$$

Add 3 to -\frac{13}{3}.

$$x=-\frac{4}{3}$$

The system is now solved.

$$x=-\frac{4}{3},y=-\frac{13}{3}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply x by 1-2x.

$$x-2x^{2}-y\left(1-y\right)=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

Use the distributive property to multiply y by 1-y.

$$x-2x^{2}-\left(y-y^{2}\right)=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

To find the opposite of y-y^{2}, find the opposite of each term.

$$x-2x^{2}-y+y^{2}=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

Consider \left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x-2x^{2}-y+y^{2}=y^{2}-\left(\sqrt{2}x\right)^{2}+3$$

Expand \left(\sqrt{2}x\right)^{2}.

$$x-2x^{2}-y+y^{2}=y^{2}-\left(\sqrt{2}\right)^{2}x^{2}+3$$

The square of \sqrt{2} is 2.

$$x-2x^{2}-y+y^{2}=y^{2}-2x^{2}+3$$

Subtract y^{2} from both sides.

$$x-2x^{2}-y+y^{2}-y^{2}=-2x^{2}+3$$

Combine y^{2} and -y^{2} to get 0.

$$x-2x^{2}-y=-2x^{2}+3$$

Add 2x^{2} to both sides.

$$x-2x^{2}-y+2x^{2}=3$$

Combine -2x^{2} and 2x^{2} to get 0.

$$x-y=3$$

Consider the second equation. Multiply both sides of the equation by 16.

$$16\left(2x-\left(2y-\frac{1}{4}\right)^{2}\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-\frac{1}{4}\right)^{2}.

$$16\left(2x-\left(4y^{2}-y+\frac{1}{16}\right)\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

To find the opposite of 4y^{2}-y+\frac{1}{16}, find the opposite of each term.

$$16\left(2x-4y^{2}+y-\frac{1}{16}\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Use the distributive property to multiply 16 by 2x-4y^{2}+y-\frac{1}{16}.

$$32x-64y^{2}+16y-1+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Multiply 16 and 16 to get 256.

$$32x-64y^{2}+16y-1+256+1=16\left(2y+3\right)\left(3-2y\right)$$

Add -1 and 256 to get 255.

$$32x-64y^{2}+16y+255+1=16\left(2y+3\right)\left(3-2y\right)$$

Add 255 and 1 to get 256.

$$32x-64y^{2}+16y+256=16\left(2y+3\right)\left(3-2y\right)$$

Use the distributive property to multiply 16 by 2y+3.

$$32x-64y^{2}+16y+256=\left(32y+48\right)\left(3-2y\right)$$

Use the distributive property to multiply 32y+48 by 3-2y and combine like terms.

$$32x-64y^{2}+16y+256=-64y^{2}+144$$

Add 64y^{2} to both sides.

$$32x-64y^{2}+16y+256+64y^{2}=144$$

Combine -64y^{2} and 64y^{2} to get 0.

$$32x+16y+256=144$$

Subtract 256 from both sides.

$$32x+16y=144-256$$

Subtract 256 from 144 to get -112.

$$32x+16y=-112$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$x-y=3,32x+16y=-112$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\32&16\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\-112\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\32&16\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\32&16\end{matrix}\right))\left(\begin{matrix}1&-1\\32&16\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\32&16\end{matrix}\right))\left(\begin{matrix}3\\-112\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\32&16\end{matrix}\right))\left(\begin{matrix}3\\-112\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\32&16\end{matrix}\right))\left(\begin{matrix}3\\-112\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{16}{16-\left(-32\right)}&-\frac{-1}{16-\left(-32\right)}\\-\frac{32}{16-\left(-32\right)}&\frac{1}{16-\left(-32\right)}\end{matrix}\right)\left(\begin{matrix}3\\-112\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}&\frac{1}{48}\\-\frac{2}{3}&\frac{1}{48}\end{matrix}\right)\left(\begin{matrix}3\\-112\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 3+\frac{1}{48}\left(-112\right)\\-\frac{2}{3}\times 3+\frac{1}{48}\left(-112\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{3}\\-\frac{13}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{4}{3},y=-\frac{13}{3}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply x by 1-2x.

$$x-2x^{2}-y\left(1-y\right)=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

Use the distributive property to multiply y by 1-y.

$$x-2x^{2}-\left(y-y^{2}\right)=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

To find the opposite of y-y^{2}, find the opposite of each term.

$$x-2x^{2}-y+y^{2}=\left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right)+3$$

Consider \left(y-\sqrt{2}x\right)\left(y+\sqrt{2}x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x-2x^{2}-y+y^{2}=y^{2}-\left(\sqrt{2}x\right)^{2}+3$$

Expand \left(\sqrt{2}x\right)^{2}.

$$x-2x^{2}-y+y^{2}=y^{2}-\left(\sqrt{2}\right)^{2}x^{2}+3$$

The square of \sqrt{2} is 2.

$$x-2x^{2}-y+y^{2}=y^{2}-2x^{2}+3$$

Subtract y^{2} from both sides.

$$x-2x^{2}-y+y^{2}-y^{2}=-2x^{2}+3$$

Combine y^{2} and -y^{2} to get 0.

$$x-2x^{2}-y=-2x^{2}+3$$

Add 2x^{2} to both sides.

$$x-2x^{2}-y+2x^{2}=3$$

Combine -2x^{2} and 2x^{2} to get 0.

$$x-y=3$$

Consider the second equation. Multiply both sides of the equation by 16.

$$16\left(2x-\left(2y-\frac{1}{4}\right)^{2}\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-\frac{1}{4}\right)^{2}.

$$16\left(2x-\left(4y^{2}-y+\frac{1}{16}\right)\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

To find the opposite of 4y^{2}-y+\frac{1}{16}, find the opposite of each term.

$$16\left(2x-4y^{2}+y-\frac{1}{16}\right)+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Use the distributive property to multiply 16 by 2x-4y^{2}+y-\frac{1}{16}.

$$32x-64y^{2}+16y-1+16\times 16+1=16\left(2y+3\right)\left(3-2y\right)$$

Multiply 16 and 16 to get 256.

$$32x-64y^{2}+16y-1+256+1=16\left(2y+3\right)\left(3-2y\right)$$

Add -1 and 256 to get 255.

$$32x-64y^{2}+16y+255+1=16\left(2y+3\right)\left(3-2y\right)$$

Add 255 and 1 to get 256.

$$32x-64y^{2}+16y+256=16\left(2y+3\right)\left(3-2y\right)$$

Use the distributive property to multiply 16 by 2y+3.

$$32x-64y^{2}+16y+256=\left(32y+48\right)\left(3-2y\right)$$

Use the distributive property to multiply 32y+48 by 3-2y and combine like terms.

$$32x-64y^{2}+16y+256=-64y^{2}+144$$

Add 64y^{2} to both sides.

$$32x-64y^{2}+16y+256+64y^{2}=144$$

Combine -64y^{2} and 64y^{2} to get 0.

$$32x+16y+256=144$$

Subtract 256 from both sides.

$$32x+16y=144-256$$

Subtract 256 from 144 to get -112.

$$32x+16y=-112$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$x-y=3,32x+16y=-112$$

To make x and 32x equal, multiply all terms on each side of the first equation by 32 and all terms on each side of the second by 1.

$$32x+32\left(-1\right)y=32\times 3,32x+16y=-112$$

Simplify.

$$32x-32y=96,32x+16y=-112$$

Subtract 32x+16y=-112 from 32x-32y=96 by subtracting like terms on each side of the equal sign.

$$32x-32x-32y-16y=96+112$$

Add 32x to -32x. Terms 32x and -32x cancel out, leaving an equation with only one variable that can be solved.

$$-32y-16y=96+112$$

Add -32y to -16y.

$$-48y=96+112$$

Add 96 to 112.

$$-48y=208$$

Divide both sides by -48.

$$y=-\frac{13}{3}$$

Substitute -\frac{13}{3} for y in 32x+16y=-112. Because the resulting equation contains only one variable, you can solve for x directly.

$$32x+16\left(-\frac{13}{3}\right)=-112$$

Multiply 16 times -\frac{13}{3}.

$$32x-\frac{208}{3}=-112$$

Add \frac{208}{3} to both sides of the equation.

$$32x=-\frac{128}{3}$$

Divide both sides by 32.

$$x=-\frac{4}{3}$$

The system is now solved.

$$x=-\frac{4}{3},y=-\frac{13}{3}$$