Solve {l}{-5x-4[x-y+(y-2)(y+2)]=-4y^2+1}{3(3x-2y-2)+x^2-3=(x-3)(x+3)+6}

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

Solve for x, y

$$x = \frac{11}{3} = 3\frac{2}{3} \approx 3.666666667$$

$$y = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Substitution

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

$$-5x-4\left(x-y+y^{2}-4\right)=-4y^{2}+1$$

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

$$-5x-4x+4y-4y^{2}+16=-4y^{2}+1$$

Combine -5x and -4x to get -9x.

$$-9x+4y-4y^{2}+16=-4y^{2}+1$$

Add 4y^{2} to both sides.

$$-9x+4y-4y^{2}+16+4y^{2}=1$$

Combine -4y^{2} and 4y^{2} to get 0.

$$-9x+4y+16=1$$

Subtract 16 from both sides.

$$-9x+4y=1-16$$

Subtract 16 from 1 to get -15.

$$-9x+4y=-15$$

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

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

Subtract 3 from -6 to get -9.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9x-6y-9+x^{2}=x^{2}-9+6$$

Add -9 and 6 to get -3.

$$9x-6y-9+x^{2}=x^{2}-3$$

Subtract x^{2} from both sides.

$$9x-6y-9+x^{2}-x^{2}=-3$$

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

$$9x-6y-9=-3$$

Add 9 to both sides.

$$9x-6y=-3+9$$

Add -3 and 9 to get 6.

$$9x-6y=6$$

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.

$$-9x+4y=-15,9x-6y=6$$

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

$$-9x+4y=-15$$

Subtract 4y from both sides of the equation.

$$-9x=-4y-15$$

Divide both sides by -9.

$$x=-\frac{1}{9}\left(-4y-15\right)$$

Multiply -\frac{1}{9} times -4y-15.

$$x=\frac{4}{9}y+\frac{5}{3}$$

Substitute \frac{4y}{9}+\frac{5}{3} for x in the other equation, 9x-6y=6.

$$9\left(\frac{4}{9}y+\frac{5}{3}\right)-6y=6$$

Multiply 9 times \frac{4y}{9}+\frac{5}{3}.

$$4y+15-6y=6$$

Add 4y to -6y.

$$-2y+15=6$$

Subtract 15 from both sides of the equation.

$$-2y=-9$$

Divide both sides by -2.

$$y=\frac{9}{2}$$

Substitute \frac{9}{2} for y in x=\frac{4}{9}y+\frac{5}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{9}\times \frac{9}{2}+\frac{5}{3}$$

Multiply \frac{4}{9} times \frac{9}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=2+\frac{5}{3}$$

Add \frac{5}{3} to 2.

$$x=\frac{11}{3}$$

The system is now solved.

$$x=\frac{11}{3},y=\frac{9}{2}$$

Steps Using Matrices

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

$$-5x-4\left(x-y+y^{2}-4\right)=-4y^{2}+1$$

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

$$-5x-4x+4y-4y^{2}+16=-4y^{2}+1$$

Combine -5x and -4x to get -9x.

$$-9x+4y-4y^{2}+16=-4y^{2}+1$$

Add 4y^{2} to both sides.

$$-9x+4y-4y^{2}+16+4y^{2}=1$$

Combine -4y^{2} and 4y^{2} to get 0.

$$-9x+4y+16=1$$

Subtract 16 from both sides.

$$-9x+4y=1-16$$

Subtract 16 from 1 to get -15.

$$-9x+4y=-15$$

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

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

Subtract 3 from -6 to get -9.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9x-6y-9+x^{2}=x^{2}-9+6$$

Add -9 and 6 to get -3.

$$9x-6y-9+x^{2}=x^{2}-3$$

Subtract x^{2} from both sides.

$$9x-6y-9+x^{2}-x^{2}=-3$$

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

$$9x-6y-9=-3$$

Add 9 to both sides.

$$9x-6y=-3+9$$

Add -3 and 9 to get 6.

$$9x-6y=6$$

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

$$-9x+4y=-15,9x-6y=6$$

Write the equations in matrix form.

$$\left(\begin{matrix}-9&4\\9&-6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-15\\6\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-9&4\\9&-6\end{matrix}\right).

$$inverse(\left(\begin{matrix}-9&4\\9&-6\end{matrix}\right))\left(\begin{matrix}-9&4\\9&-6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-9&4\\9&-6\end{matrix}\right))\left(\begin{matrix}-15\\6\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-9&4\\9&-6\end{matrix}\right))\left(\begin{matrix}-15\\6\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-9&4\\9&-6\end{matrix}\right))\left(\begin{matrix}-15\\6\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{-9\left(-6\right)-4\times 9}&-\frac{4}{-9\left(-6\right)-4\times 9}\\-\frac{9}{-9\left(-6\right)-4\times 9}&-\frac{9}{-9\left(-6\right)-4\times 9}\end{matrix}\right)\left(\begin{matrix}-15\\6\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}&-\frac{2}{9}\\-\frac{1}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}-15\\6\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\left(-15\right)-\frac{2}{9}\times 6\\-\frac{1}{2}\left(-15\right)-\frac{1}{2}\times 6\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{3}\\\frac{9}{2}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{11}{3},y=\frac{9}{2}$$

Steps Using Elimination

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

$$-5x-4\left(x-y+y^{2}-4\right)=-4y^{2}+1$$

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

$$-5x-4x+4y-4y^{2}+16=-4y^{2}+1$$

Combine -5x and -4x to get -9x.

$$-9x+4y-4y^{2}+16=-4y^{2}+1$$

Add 4y^{2} to both sides.

$$-9x+4y-4y^{2}+16+4y^{2}=1$$

Combine -4y^{2} and 4y^{2} to get 0.

$$-9x+4y+16=1$$

Subtract 16 from both sides.

$$-9x+4y=1-16$$

Subtract 16 from 1 to get -15.

$$-9x+4y=-15$$

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

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

Subtract 3 from -6 to get -9.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9x-6y-9+x^{2}=x^{2}-9+6$$

Add -9 and 6 to get -3.

$$9x-6y-9+x^{2}=x^{2}-3$$

Subtract x^{2} from both sides.

$$9x-6y-9+x^{2}-x^{2}=-3$$

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

$$9x-6y-9=-3$$

Add 9 to both sides.

$$9x-6y=-3+9$$

Add -3 and 9 to get 6.

$$9x-6y=6$$

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.

$$-9x+4y=-15,9x-6y=6$$

To make -9x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by -9.

$$9\left(-9\right)x+9\times 4y=9\left(-15\right),-9\times 9x-9\left(-6\right)y=-9\times 6$$

Simplify.

$$-81x+36y=-135,-81x+54y=-54$$

Subtract -81x+54y=-54 from -81x+36y=-135 by subtracting like terms on each side of the equal sign.

$$-81x+81x+36y-54y=-135+54$$

Add -81x to 81x. Terms -81x and 81x cancel out, leaving an equation with only one variable that can be solved.

$$36y-54y=-135+54$$

Add 36y to -54y.

$$-18y=-135+54$$

Add -135 to 54.

$$-18y=-81$$

Divide both sides by -18.

$$y=\frac{9}{2}$$

Substitute \frac{9}{2} for y in 9x-6y=6. Because the resulting equation contains only one variable, you can solve for x directly.

$$9x-6\times \frac{9}{2}=6$$

Multiply -6 times \frac{9}{2}.

$$9x-27=6$$

Add 27 to both sides of the equation.

$$9x=33$$

Divide both sides by 9.

$$x=\frac{11}{3}$$

The system is now solved.

$$x=\frac{11}{3},y=\frac{9}{2}$$