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

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

Solve for x, y

$$x=0$$

$$y=1$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\times 2\left(x-1\right)+12y=3x+8$$

Multiply 2 and 2 to get 4.

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

Use the distributive property to multiply 4 by x-1.

$$4x-4+12y=3x+8$$

Subtract 3x from both sides.

$$4x-4+12y-3x=8$$

Combine 4x and -3x to get x.

$$x-4+12y=8$$

Add 4 to both sides.

$$x+12y=8+4$$

Add 8 and 4 to get 12.

$$x+12y=12$$

Consider the second equation. Multiply both sides of the equation by 3.

$$4x-3\left(y+2\right)=-9$$

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

$$4x-3y-6=-9$$

Add 6 to both sides.

$$4x-3y=-9+6$$

Add -9 and 6 to get -3.

$$4x-3y=-3$$

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.

$$x+12y=12,4x-3y=-3$$

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

$$x+12y=12$$

Subtract 12y from both sides of the equation.

$$x=-12y+12$$

Substitute -12y+12 for x in the other equation, 4x-3y=-3.

$$4\left(-12y+12\right)-3y=-3$$

Multiply 4 times -12y+12.

$$-48y+48-3y=-3$$

Add -48y to -3y.

$$-51y+48=-3$$

Subtract 48 from both sides of the equation.

$$-51y=-51$$

Divide both sides by -51.

$$y=1$$

Substitute 1 for y in x=-12y+12. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-12+12$$

Add 12 to -12.

$$x=0$$

The system is now solved.

$$x=0,y=1$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\times 2\left(x-1\right)+12y=3x+8$$

Multiply 2 and 2 to get 4.

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

Use the distributive property to multiply 4 by x-1.

$$4x-4+12y=3x+8$$

Subtract 3x from both sides.

$$4x-4+12y-3x=8$$

Combine 4x and -3x to get x.

$$x-4+12y=8$$

Add 4 to both sides.

$$x+12y=8+4$$

Add 8 and 4 to get 12.

$$x+12y=12$$

Consider the second equation. Multiply both sides of the equation by 3.

$$4x-3\left(y+2\right)=-9$$

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

$$4x-3y-6=-9$$

Add 6 to both sides.

$$4x-3y=-9+6$$

Add -9 and 6 to get -3.

$$4x-3y=-3$$

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

$$x+12y=12,4x-3y=-3$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&12\\4&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\-3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&12\\4&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&12\\4&-3\end{matrix}\right))\left(\begin{matrix}1&12\\4&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&12\\4&-3\end{matrix}\right))\left(\begin{matrix}12\\-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&12\\4&-3\end{matrix}\right))\left(\begin{matrix}12\\-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&12\\4&-3\end{matrix}\right))\left(\begin{matrix}12\\-3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-12\times 4}&-\frac{12}{-3-12\times 4}\\-\frac{4}{-3-12\times 4}&\frac{1}{-3-12\times 4}\end{matrix}\right)\left(\begin{matrix}12\\-3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{17}&\frac{4}{17}\\\frac{4}{51}&-\frac{1}{51}\end{matrix}\right)\left(\begin{matrix}12\\-3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{17}\times 12+\frac{4}{17}\left(-3\right)\\\frac{4}{51}\times 12-\frac{1}{51}\left(-3\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=1$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\times 2\left(x-1\right)+12y=3x+8$$

Multiply 2 and 2 to get 4.

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

Use the distributive property to multiply 4 by x-1.

$$4x-4+12y=3x+8$$

Subtract 3x from both sides.

$$4x-4+12y-3x=8$$

Combine 4x and -3x to get x.

$$x-4+12y=8$$

Add 4 to both sides.

$$x+12y=8+4$$

Add 8 and 4 to get 12.

$$x+12y=12$$

Consider the second equation. Multiply both sides of the equation by 3.

$$4x-3\left(y+2\right)=-9$$

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

$$4x-3y-6=-9$$

Add 6 to both sides.

$$4x-3y=-9+6$$

Add -9 and 6 to get -3.

$$4x-3y=-3$$

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.

$$x+12y=12,4x-3y=-3$$

To make x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.

$$4x+4\times 12y=4\times 12,4x-3y=-3$$

Simplify.

$$4x+48y=48,4x-3y=-3$$

Subtract 4x-3y=-3 from 4x+48y=48 by subtracting like terms on each side of the equal sign.

$$4x-4x+48y+3y=48+3$$

Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.

$$48y+3y=48+3$$

Add 48y to 3y.

$$51y=48+3$$

Add 48 to 3.

$$51y=51$$

Divide both sides by 51.

$$y=1$$

Substitute 1 for y in 4x-3y=-3. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x-3=-3$$

Add 3 to both sides of the equation.

$$4x=0$$

Divide both sides by 4.

$$x=0$$

The system is now solved.

$$x=0,y=1$$