Solve {l}{(2x+1)[1-6x+(2x+1)^2]=2x(2x+3)(2x-3)+2y}{(3x+2)^2-(3x+1)^2-3(y+8)=0}

$$\left. \begin{array} { l } { ( 2 x + 1 ) [ 1 - 6 x + ( 2 x + 1 ) ^ { 2 } ] = 2 x ( 2 x + 3 ) ( 2 x - 3 ) + 2 y } \\ { ( 3 x + 2 ) ^ { 2 } - ( 3 x + 1 ) ^ { 2 } - 3 ( y + 8 ) = 0 } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=-9$$

Steps Using Substitution

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$\left(2x+1\right)\left(1-6x+4x^{2}+4x+1\right)=2x\left(2x+3\right)\left(2x-3\right)+2y$$

Combine -6x and 4x to get -2x.

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

Add 1 and 1 to get 2.

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

Use the distributive property to multiply 2x+1 by 2-2x+4x^{2} and combine like terms.

$$2x+8x^{3}+2=2x\left(2x+3\right)\left(2x-3\right)+2y$$

Use the distributive property to multiply 2x by 2x+3.

$$2x+8x^{3}+2=\left(4x^{2}+6x\right)\left(2x-3\right)+2y$$

Use the distributive property to multiply 4x^{2}+6x by 2x-3 and combine like terms.

$$2x+8x^{3}+2=8x^{3}-18x+2y$$

Subtract 8x^{3} from both sides.

$$2x+8x^{3}+2-8x^{3}=-18x+2y$$

Combine 8x^{3} and -8x^{3} to get 0.

$$2x+2=-18x+2y$$

Add 18x to both sides.

$$2x+2+18x=2y$$

Combine 2x and 18x to get 20x.

$$20x+2=2y$$

Subtract 2y from both sides.

$$20x+2-2y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$20x-2y=-2$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+12x+4-\left(3x+1\right)^{2}-3\left(y+8\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$9x^{2}+12x+4-\left(9x^{2}+6x+1\right)-3\left(y+8\right)=0$$

To find the opposite of 9x^{2}+6x+1, find the opposite of each term.

$$9x^{2}+12x+4-9x^{2}-6x-1-3\left(y+8\right)=0$$

Combine 9x^{2} and -9x^{2} to get 0.

$$12x+4-6x-1-3\left(y+8\right)=0$$

Combine 12x and -6x to get 6x.

$$6x+4-1-3\left(y+8\right)=0$$

Subtract 1 from 4 to get 3.

$$6x+3-3\left(y+8\right)=0$$

Use the distributive property to multiply -3 by y+8.

$$6x+3-3y-24=0$$

Subtract 24 from 3 to get -21.

$$6x-21-3y=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$6x-3y=21$$

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.

$$20x-2y=-2,6x-3y=21$$

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

$$20x-2y=-2$$

Add 2y to both sides of the equation.

$$20x=2y-2$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(2y-2\right)$$

Multiply \frac{1}{20} times -2+2y.

$$x=\frac{1}{10}y-\frac{1}{10}$$

Substitute \frac{-1+y}{10} for x in the other equation, 6x-3y=21.

$$6\left(\frac{1}{10}y-\frac{1}{10}\right)-3y=21$$

Multiply 6 times \frac{-1+y}{10}.

$$\frac{3}{5}y-\frac{3}{5}-3y=21$$

Add \frac{3y}{5} to -3y.

$$-\frac{12}{5}y-\frac{3}{5}=21$$

Add \frac{3}{5} to both sides of the equation.

$$-\frac{12}{5}y=\frac{108}{5}$$

Divide both sides of the equation by -\frac{12}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-9$$

Substitute -9 for y in x=\frac{1}{10}y-\frac{1}{10}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{10}\left(-9\right)-\frac{1}{10}$$

Multiply \frac{1}{10} times -9.

$$x=\frac{-9-1}{10}$$

Add -\frac{1}{10} to -\frac{9}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-1$$

The system is now solved.

$$x=-1,y=-9$$

Steps Using Matrices

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$\left(2x+1\right)\left(1-6x+4x^{2}+4x+1\right)=2x\left(2x+3\right)\left(2x-3\right)+2y$$

Combine -6x and 4x to get -2x.

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

Add 1 and 1 to get 2.

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

Use the distributive property to multiply 2x+1 by 2-2x+4x^{2} and combine like terms.

$$2x+8x^{3}+2=2x\left(2x+3\right)\left(2x-3\right)+2y$$

Use the distributive property to multiply 2x by 2x+3.

$$2x+8x^{3}+2=\left(4x^{2}+6x\right)\left(2x-3\right)+2y$$

Use the distributive property to multiply 4x^{2}+6x by 2x-3 and combine like terms.

$$2x+8x^{3}+2=8x^{3}-18x+2y$$

Subtract 8x^{3} from both sides.

$$2x+8x^{3}+2-8x^{3}=-18x+2y$$

Combine 8x^{3} and -8x^{3} to get 0.

$$2x+2=-18x+2y$$

Add 18x to both sides.

$$2x+2+18x=2y$$

Combine 2x and 18x to get 20x.

$$20x+2=2y$$

Subtract 2y from both sides.

$$20x+2-2y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$20x-2y=-2$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+12x+4-\left(3x+1\right)^{2}-3\left(y+8\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$9x^{2}+12x+4-\left(9x^{2}+6x+1\right)-3\left(y+8\right)=0$$

To find the opposite of 9x^{2}+6x+1, find the opposite of each term.

$$9x^{2}+12x+4-9x^{2}-6x-1-3\left(y+8\right)=0$$

Combine 9x^{2} and -9x^{2} to get 0.

$$12x+4-6x-1-3\left(y+8\right)=0$$

Combine 12x and -6x to get 6x.

$$6x+4-1-3\left(y+8\right)=0$$

Subtract 1 from 4 to get 3.

$$6x+3-3\left(y+8\right)=0$$

Use the distributive property to multiply -3 by y+8.

$$6x+3-3y-24=0$$

Subtract 24 from 3 to get -21.

$$6x-21-3y=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$6x-3y=21$$

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

$$20x-2y=-2,6x-3y=21$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&-2\\6&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\\21\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&-2\\6&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&-2\\6&-3\end{matrix}\right))\left(\begin{matrix}20&-2\\6&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-2\\6&-3\end{matrix}\right))\left(\begin{matrix}-2\\21\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-2\\6&-3\end{matrix}\right))\left(\begin{matrix}-2\\21\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-2\\6&-3\end{matrix}\right))\left(\begin{matrix}-2\\21\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{20\left(-3\right)-\left(-2\times 6\right)}&-\frac{-2}{20\left(-3\right)-\left(-2\times 6\right)}\\-\frac{6}{20\left(-3\right)-\left(-2\times 6\right)}&\frac{20}{20\left(-3\right)-\left(-2\times 6\right)}\end{matrix}\right)\left(\begin{matrix}-2\\21\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{16}&-\frac{1}{24}\\\frac{1}{8}&-\frac{5}{12}\end{matrix}\right)\left(\begin{matrix}-2\\21\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{16}\left(-2\right)-\frac{1}{24}\times 21\\\frac{1}{8}\left(-2\right)-\frac{5}{12}\times 21\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\-9\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-1,y=-9$$

Steps Using Elimination

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$\left(2x+1\right)\left(1-6x+4x^{2}+4x+1\right)=2x\left(2x+3\right)\left(2x-3\right)+2y$$

Combine -6x and 4x to get -2x.

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

Add 1 and 1 to get 2.

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

Use the distributive property to multiply 2x+1 by 2-2x+4x^{2} and combine like terms.

$$2x+8x^{3}+2=2x\left(2x+3\right)\left(2x-3\right)+2y$$

Use the distributive property to multiply 2x by 2x+3.

$$2x+8x^{3}+2=\left(4x^{2}+6x\right)\left(2x-3\right)+2y$$

Use the distributive property to multiply 4x^{2}+6x by 2x-3 and combine like terms.

$$2x+8x^{3}+2=8x^{3}-18x+2y$$

Subtract 8x^{3} from both sides.

$$2x+8x^{3}+2-8x^{3}=-18x+2y$$

Combine 8x^{3} and -8x^{3} to get 0.

$$2x+2=-18x+2y$$

Add 18x to both sides.

$$2x+2+18x=2y$$

Combine 2x and 18x to get 20x.

$$20x+2=2y$$

Subtract 2y from both sides.

$$20x+2-2y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$20x-2y=-2$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+12x+4-\left(3x+1\right)^{2}-3\left(y+8\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$9x^{2}+12x+4-\left(9x^{2}+6x+1\right)-3\left(y+8\right)=0$$

To find the opposite of 9x^{2}+6x+1, find the opposite of each term.

$$9x^{2}+12x+4-9x^{2}-6x-1-3\left(y+8\right)=0$$

Combine 9x^{2} and -9x^{2} to get 0.

$$12x+4-6x-1-3\left(y+8\right)=0$$

Combine 12x and -6x to get 6x.

$$6x+4-1-3\left(y+8\right)=0$$

Subtract 1 from 4 to get 3.

$$6x+3-3\left(y+8\right)=0$$

Use the distributive property to multiply -3 by y+8.

$$6x+3-3y-24=0$$

Subtract 24 from 3 to get -21.

$$6x-21-3y=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$6x-3y=21$$

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.

$$20x-2y=-2,6x-3y=21$$

To make 20x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 20.

$$6\times 20x+6\left(-2\right)y=6\left(-2\right),20\times 6x+20\left(-3\right)y=20\times 21$$

Simplify.

$$120x-12y=-12,120x-60y=420$$

Subtract 120x-60y=420 from 120x-12y=-12 by subtracting like terms on each side of the equal sign.

$$120x-120x-12y+60y=-12-420$$

Add 120x to -120x. Terms 120x and -120x cancel out, leaving an equation with only one variable that can be solved.

$$-12y+60y=-12-420$$

Add -12y to 60y.

$$48y=-12-420$$

Add -12 to -420.

$$48y=-432$$

Divide both sides by 48.

$$y=-9$$

Substitute -9 for y in 6x-3y=21. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x-3\left(-9\right)=21$$

Multiply -3 times -9.

$$6x+27=21$$

Subtract 27 from both sides of the equation.

$$6x=-6$$

Divide both sides by 6.

$$x=-1$$

The system is now solved.

$$x=-1,y=-9$$