Solve {l}{(3x-1)^{2}(2x+3)-8y=3x^{2}(6x+5)+y-x}{y(2y-3)(3-2y)+16x^2-54=(4x-3)^2-4(y^3-3y^2-3)}

$$\left\{ \begin{array} { l } { ( 3 x - 1 ) ^ { 2 } ( 2 x + 3 ) - 8 y = 3 x ^ { 2 } ( 6 x + 5 ) + y - x } \\ { y ( 2 y - 3 ) ( 3 - 2 y ) + 16 x ^ { 2 } - 54 = ( 4 x - 3 ) ^ { 2 } - 4 ( y ^ { 3 } - 3 y ^ { 2 } - 3 ) } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=-3$$

Steps Using Substitution

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

$$\left(9x^{2}-6x+1\right)\left(2x+3\right)-8y=3x^{2}\left(6x+5\right)+y-x$$

Use the distributive property to multiply 9x^{2}-6x+1 by 2x+3 and combine like terms.

$$18x^{3}+15x^{2}-16x+3-8y=3x^{2}\left(6x+5\right)+y-x$$

Use the distributive property to multiply 3x^{2} by 6x+5.

$$18x^{3}+15x^{2}-16x+3-8y=18x^{3}+15x^{2}+y-x$$

Subtract 18x^{3} from both sides.

$$18x^{3}+15x^{2}-16x+3-8y-18x^{3}=15x^{2}+y-x$$

Combine 18x^{3} and -18x^{3} to get 0.

$$15x^{2}-16x+3-8y=15x^{2}+y-x$$

Subtract 15x^{2} from both sides.

$$15x^{2}-16x+3-8y-15x^{2}=y-x$$

Combine 15x^{2} and -15x^{2} to get 0.

$$-16x+3-8y=y-x$$

Subtract y from both sides.

$$-16x+3-8y-y=-x$$

Combine -8y and -y to get -9y.

$$-16x+3-9y=-x$$

Add x to both sides.

$$-16x+3-9y+x=0$$

Combine -16x and x to get -15x.

$$-15x+3-9y=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-15x-9y=-3$$

Consider the second equation. Use the distributive property to multiply y by 2y-3.

$$\left(2y^{2}-3y\right)\left(3-2y\right)+16x^{2}-54=\left(4x-3\right)^{2}-4\left(y^{3}-3y^{2}-3\right)$$

Use the distributive property to multiply 2y^{2}-3y by 3-2y and combine like terms.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=\left(4x-3\right)^{2}-4\left(y^{3}-3y^{2}-3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+9-4\left(y^{3}-3y^{2}-3\right)$$

Use the distributive property to multiply -4 by y^{3}-3y^{2}-3.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+9-4y^{3}+12y^{2}+12$$

Add 9 and 12 to get 21.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+21-4y^{3}+12y^{2}$$

Subtract 16x^{2} from both sides.

$$12y^{2}-4y^{3}-9y+16x^{2}-54-16x^{2}=-24x+21-4y^{3}+12y^{2}$$

Combine 16x^{2} and -16x^{2} to get 0.

$$12y^{2}-4y^{3}-9y-54=-24x+21-4y^{3}+12y^{2}$$

Add 24x to both sides.

$$12y^{2}-4y^{3}-9y-54+24x=21-4y^{3}+12y^{2}$$

Add 4y^{3} to both sides.

$$12y^{2}-4y^{3}-9y-54+24x+4y^{3}=21+12y^{2}$$

Combine -4y^{3} and 4y^{3} to get 0.

$$12y^{2}-9y-54+24x=21+12y^{2}$$

Subtract 12y^{2} from both sides.

$$12y^{2}-9y-54+24x-12y^{2}=21$$

Combine 12y^{2} and -12y^{2} to get 0.

$$-9y-54+24x=21$$

Add 54 to both sides.

$$-9y+24x=21+54$$

Add 21 and 54 to get 75.

$$-9y+24x=75$$

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.

$$-15x-9y=-3,24x-9y=75$$

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

$$-15x-9y=-3$$

Add 9y to both sides of the equation.

$$-15x=9y-3$$

Divide both sides by -15.

$$x=-\frac{1}{15}\left(9y-3\right)$$

Multiply -\frac{1}{15} times 9y-3.

$$x=-\frac{3}{5}y+\frac{1}{5}$$

Substitute \frac{-3y+1}{5} for x in the other equation, 24x-9y=75.

$$24\left(-\frac{3}{5}y+\frac{1}{5}\right)-9y=75$$

Multiply 24 times \frac{-3y+1}{5}.

$$-\frac{72}{5}y+\frac{24}{5}-9y=75$$

Add -\frac{72y}{5} to -9y.

$$-\frac{117}{5}y+\frac{24}{5}=75$$

Subtract \frac{24}{5} from both sides of the equation.

$$-\frac{117}{5}y=\frac{351}{5}$$

Divide both sides of the equation by -\frac{117}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-3$$

Substitute -3 for y in x=-\frac{3}{5}y+\frac{1}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{5}\left(-3\right)+\frac{1}{5}$$

Multiply -\frac{3}{5} times -3.

$$x=\frac{9+1}{5}$$

Add \frac{1}{5} to \frac{9}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=2$$

The system is now solved.

$$x=2,y=-3$$

Steps Using Matrices

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

$$\left(9x^{2}-6x+1\right)\left(2x+3\right)-8y=3x^{2}\left(6x+5\right)+y-x$$

Use the distributive property to multiply 9x^{2}-6x+1 by 2x+3 and combine like terms.

$$18x^{3}+15x^{2}-16x+3-8y=3x^{2}\left(6x+5\right)+y-x$$

Use the distributive property to multiply 3x^{2} by 6x+5.

$$18x^{3}+15x^{2}-16x+3-8y=18x^{3}+15x^{2}+y-x$$

Subtract 18x^{3} from both sides.

$$18x^{3}+15x^{2}-16x+3-8y-18x^{3}=15x^{2}+y-x$$

Combine 18x^{3} and -18x^{3} to get 0.

$$15x^{2}-16x+3-8y=15x^{2}+y-x$$

Subtract 15x^{2} from both sides.

$$15x^{2}-16x+3-8y-15x^{2}=y-x$$

Combine 15x^{2} and -15x^{2} to get 0.

$$-16x+3-8y=y-x$$

Subtract y from both sides.

$$-16x+3-8y-y=-x$$

Combine -8y and -y to get -9y.

$$-16x+3-9y=-x$$

Add x to both sides.

$$-16x+3-9y+x=0$$

Combine -16x and x to get -15x.

$$-15x+3-9y=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-15x-9y=-3$$

Consider the second equation. Use the distributive property to multiply y by 2y-3.

$$\left(2y^{2}-3y\right)\left(3-2y\right)+16x^{2}-54=\left(4x-3\right)^{2}-4\left(y^{3}-3y^{2}-3\right)$$

Use the distributive property to multiply 2y^{2}-3y by 3-2y and combine like terms.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=\left(4x-3\right)^{2}-4\left(y^{3}-3y^{2}-3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+9-4\left(y^{3}-3y^{2}-3\right)$$

Use the distributive property to multiply -4 by y^{3}-3y^{2}-3.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+9-4y^{3}+12y^{2}+12$$

Add 9 and 12 to get 21.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+21-4y^{3}+12y^{2}$$

Subtract 16x^{2} from both sides.

$$12y^{2}-4y^{3}-9y+16x^{2}-54-16x^{2}=-24x+21-4y^{3}+12y^{2}$$

Combine 16x^{2} and -16x^{2} to get 0.

$$12y^{2}-4y^{3}-9y-54=-24x+21-4y^{3}+12y^{2}$$

Add 24x to both sides.

$$12y^{2}-4y^{3}-9y-54+24x=21-4y^{3}+12y^{2}$$

Add 4y^{3} to both sides.

$$12y^{2}-4y^{3}-9y-54+24x+4y^{3}=21+12y^{2}$$

Combine -4y^{3} and 4y^{3} to get 0.

$$12y^{2}-9y-54+24x=21+12y^{2}$$

Subtract 12y^{2} from both sides.

$$12y^{2}-9y-54+24x-12y^{2}=21$$

Combine 12y^{2} and -12y^{2} to get 0.

$$-9y-54+24x=21$$

Add 54 to both sides.

$$-9y+24x=21+54$$

Add 21 and 54 to get 75.

$$-9y+24x=75$$

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

$$-15x-9y=-3,24x-9y=75$$

Write the equations in matrix form.

$$\left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\\75\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right).

$$inverse(\left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right))\left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right))\left(\begin{matrix}-3\\75\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right))\left(\begin{matrix}-3\\75\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-15&-9\\24&-9\end{matrix}\right))\left(\begin{matrix}-3\\75\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{-15\left(-9\right)-\left(-9\times 24\right)}&-\frac{-9}{-15\left(-9\right)-\left(-9\times 24\right)}\\-\frac{24}{-15\left(-9\right)-\left(-9\times 24\right)}&-\frac{15}{-15\left(-9\right)-\left(-9\times 24\right)}\end{matrix}\right)\left(\begin{matrix}-3\\75\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{39}&\frac{1}{39}\\-\frac{8}{117}&-\frac{5}{117}\end{matrix}\right)\left(\begin{matrix}-3\\75\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{39}\left(-3\right)+\frac{1}{39}\times 75\\-\frac{8}{117}\left(-3\right)-\frac{5}{117}\times 75\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=2,y=-3$$

Steps Using Elimination

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

$$\left(9x^{2}-6x+1\right)\left(2x+3\right)-8y=3x^{2}\left(6x+5\right)+y-x$$

Use the distributive property to multiply 9x^{2}-6x+1 by 2x+3 and combine like terms.

$$18x^{3}+15x^{2}-16x+3-8y=3x^{2}\left(6x+5\right)+y-x$$

Use the distributive property to multiply 3x^{2} by 6x+5.

$$18x^{3}+15x^{2}-16x+3-8y=18x^{3}+15x^{2}+y-x$$

Subtract 18x^{3} from both sides.

$$18x^{3}+15x^{2}-16x+3-8y-18x^{3}=15x^{2}+y-x$$

Combine 18x^{3} and -18x^{3} to get 0.

$$15x^{2}-16x+3-8y=15x^{2}+y-x$$

Subtract 15x^{2} from both sides.

$$15x^{2}-16x+3-8y-15x^{2}=y-x$$

Combine 15x^{2} and -15x^{2} to get 0.

$$-16x+3-8y=y-x$$

Subtract y from both sides.

$$-16x+3-8y-y=-x$$

Combine -8y and -y to get -9y.

$$-16x+3-9y=-x$$

Add x to both sides.

$$-16x+3-9y+x=0$$

Combine -16x and x to get -15x.

$$-15x+3-9y=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-15x-9y=-3$$

Consider the second equation. Use the distributive property to multiply y by 2y-3.

$$\left(2y^{2}-3y\right)\left(3-2y\right)+16x^{2}-54=\left(4x-3\right)^{2}-4\left(y^{3}-3y^{2}-3\right)$$

Use the distributive property to multiply 2y^{2}-3y by 3-2y and combine like terms.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=\left(4x-3\right)^{2}-4\left(y^{3}-3y^{2}-3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+9-4\left(y^{3}-3y^{2}-3\right)$$

Use the distributive property to multiply -4 by y^{3}-3y^{2}-3.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+9-4y^{3}+12y^{2}+12$$

Add 9 and 12 to get 21.

$$12y^{2}-4y^{3}-9y+16x^{2}-54=16x^{2}-24x+21-4y^{3}+12y^{2}$$

Subtract 16x^{2} from both sides.

$$12y^{2}-4y^{3}-9y+16x^{2}-54-16x^{2}=-24x+21-4y^{3}+12y^{2}$$

Combine 16x^{2} and -16x^{2} to get 0.

$$12y^{2}-4y^{3}-9y-54=-24x+21-4y^{3}+12y^{2}$$

Add 24x to both sides.

$$12y^{2}-4y^{3}-9y-54+24x=21-4y^{3}+12y^{2}$$

Add 4y^{3} to both sides.

$$12y^{2}-4y^{3}-9y-54+24x+4y^{3}=21+12y^{2}$$

Combine -4y^{3} and 4y^{3} to get 0.

$$12y^{2}-9y-54+24x=21+12y^{2}$$

Subtract 12y^{2} from both sides.

$$12y^{2}-9y-54+24x-12y^{2}=21$$

Combine 12y^{2} and -12y^{2} to get 0.

$$-9y-54+24x=21$$

Add 54 to both sides.

$$-9y+24x=21+54$$

Add 21 and 54 to get 75.

$$-9y+24x=75$$

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.

$$-15x-9y=-3,24x-9y=75$$

Subtract 24x-9y=75 from -15x-9y=-3 by subtracting like terms on each side of the equal sign.

$$-15x-24x-9y+9y=-3-75$$

Add -9y to 9y. Terms -9y and 9y cancel out, leaving an equation with only one variable that can be solved.

$$-15x-24x=-3-75$$

Add -15x to -24x.

$$-39x=-3-75$$

Add -3 to -75.

$$-39x=-78$$

Divide both sides by -39.

$$x=2$$

Substitute 2 for x in 24x-9y=75. Because the resulting equation contains only one variable, you can solve for y directly.

$$24\times 2-9y=75$$

Multiply 24 times 2.

$$48-9y=75$$

Subtract 48 from both sides of the equation.

$$-9y=27$$

Divide both sides by -9.

$$y=-3$$

The system is now solved.

$$x=2,y=-3$$