Solve {l}{2x+3(y-6)=0}{3x-4(2-y)=19}

$$\left. \begin{array} { l } { 2 x + 3 ( y - 6 ) = 0 } \\ { 3 x - 4 ( 2 - y ) = 19 } \end{array} \right.$$

Solve for x, y

$$x=9$$

$$y=0$$

Steps Using Substitution

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.

$$2x+3\left(y-6\right)=0,3x-4\left(-y+2\right)=19$$

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

$$2x+3\left(y-6\right)=0$$

Multiply 3 times y-6.

$$2x+3y-18=0$$

Add 18 to both sides of the equation.

$$2x+3y=18$$

Subtract 3y from both sides of the equation.

$$2x=-3y+18$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-3y+18\right)$$

Multiply \frac{1}{2} times -3y+18.

$$x=-\frac{3}{2}y+9$$

Substitute -\frac{3y}{2}+9 for x in the other equation, 3x-4\left(-y+2\right)=19.

$$3\left(-\frac{3}{2}y+9\right)-4\left(-y+2\right)=19$$

Multiply 3 times -\frac{3y}{2}+9.

$$-\frac{9}{2}y+27-4\left(-y+2\right)=19$$

Multiply -4 times -y+2.

$$-\frac{9}{2}y+27+4y-8=19$$

Add -\frac{9y}{2} to 4y.

$$-\frac{1}{2}y+27-8=19$$

Add 27 to -8.

$$-\frac{1}{2}y+19=19$$

Subtract 19 from both sides of the equation.

$$-\frac{1}{2}y=0$$

Multiply both sides by -2.

$$y=0$$

Substitute 0 for y in x=-\frac{3}{2}y+9. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=9$$

The system is now solved.

$$x=9,y=0$$

Steps Using Matrices

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

$$2x+3\left(y-6\right)=0,3x-4\left(-y+2\right)=19$$

Simplify the first equation to put it in standard form.

$$2x+3\left(y-6\right)=0$$

Multiply 3 times y-6.

$$2x+3y-18=0$$

Add 18 to both sides of the equation.

$$2x+3y=18$$

Simplify the second equation to put it in standard form.

$$3x-4\left(-y+2\right)=19$$

Multiply -4 times -y+2.

$$3x+4y-8=19$$

Add 8 to both sides of the equation.

$$3x+4y=27$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&3\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18\\27\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&3\\3&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}2&3\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}18\\27\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}18\\27\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&4\end{matrix}\right))\left(\begin{matrix}18\\27\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{2\times 4-3\times 3}&-\frac{3}{2\times 4-3\times 3}\\-\frac{3}{2\times 4-3\times 3}&\frac{2}{2\times 4-3\times 3}\end{matrix}\right)\left(\begin{matrix}18\\27\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4&3\\3&-2\end{matrix}\right)\left(\begin{matrix}18\\27\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\times 18+3\times 27\\3\times 18-2\times 27\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=9,y=0$$