Solve {l}{2(x-y)/3-x+y/4=-1D}{6(x+y)-4(2x-y)=16(2)}

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

Solve for x, y

$$x=\frac{88-30D}{7}$$

$$y=\frac{40-6D}{7}$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2\left(x-y\right)-3\left(x+y\right)=-12D$$

Multiply 4 and 2 to get 8.

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

Use the distributive property to multiply 8 by x-y.

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

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

$$8x-8y-3x-3y=-12D$$

Combine 8x and -3x to get 5x.

$$5x-8y-3y=-12D$$

Combine -8y and -3y to get -11y.

$$5x-11y=-12D$$

Consider the second equation. Use the distributive property to multiply 6 by x+y.

$$6x+6y-4\left(2x-y\right)=16\times 2$$

Use the distributive property to multiply -4 by 2x-y.

$$6x+6y-8x+4y=16\times 2$$

Combine 6x and -8x to get -2x.

$$-2x+6y+4y=16\times 2$$

Combine 6y and 4y to get 10y.

$$-2x+10y=16\times 2$$

Multiply 16 and 2 to get 32.

$$-2x+10y=32$$

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.

$$5x-11y=-12D,-2x+10y=32$$

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

$$5x-11y=-12D$$

Add 11y to both sides of the equation.

$$5x=11y-12D$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(11y-12D\right)$$

Multiply \frac{1}{5} times 11y-12D.

$$x=\frac{11}{5}y-\frac{12D}{5}$$

Substitute \frac{11y-12D}{5} for x in the other equation, -2x+10y=32.

$$-2\left(\frac{11}{5}y-\frac{12D}{5}\right)+10y=32$$

Multiply -2 times \frac{11y-12D}{5}.

$$-\frac{22}{5}y+\frac{24D}{5}+10y=32$$

Add -\frac{22y}{5} to 10y.

$$\frac{28}{5}y+\frac{24D}{5}=32$$

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

$$\frac{28}{5}y=-\frac{24D}{5}+32$$

Divide both sides of the equation by \frac{28}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{40-6D}{7}$$

Substitute \frac{40-6D}{7} for y in x=\frac{11}{5}y-\frac{12D}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{11}{5}\times \frac{40-6D}{7}-\frac{12D}{5}$$

Multiply \frac{11}{5} times \frac{40-6D}{7}.

$$x=-\frac{66D}{35}+\frac{88}{7}-\frac{12D}{5}$$

Add -\frac{12D}{5} to \frac{88}{7}-\frac{66D}{35}.

$$x=\frac{88-30D}{7}$$

The system is now solved.

$$x=\frac{88-30D}{7},y=\frac{40-6D}{7}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2\left(x-y\right)-3\left(x+y\right)=-12D$$

Multiply 4 and 2 to get 8.

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

Use the distributive property to multiply 8 by x-y.

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

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

$$8x-8y-3x-3y=-12D$$

Combine 8x and -3x to get 5x.

$$5x-8y-3y=-12D$$

Combine -8y and -3y to get -11y.

$$5x-11y=-12D$$

Consider the second equation. Use the distributive property to multiply 6 by x+y.

$$6x+6y-4\left(2x-y\right)=16\times 2$$

Use the distributive property to multiply -4 by 2x-y.

$$6x+6y-8x+4y=16\times 2$$

Combine 6x and -8x to get -2x.

$$-2x+6y+4y=16\times 2$$

Combine 6y and 4y to get 10y.

$$-2x+10y=16\times 2$$

Multiply 16 and 2 to get 32.

$$-2x+10y=32$$

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

$$5x-11y=-12D,-2x+10y=32$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-11\\-2&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-12D\\32\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-11\\-2&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-11\\-2&10\end{matrix}\right))\left(\begin{matrix}5&-11\\-2&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-11\\-2&10\end{matrix}\right))\left(\begin{matrix}-12D\\32\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-11\\-2&10\end{matrix}\right))\left(\begin{matrix}-12D\\32\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-11\\-2&10\end{matrix}\right))\left(\begin{matrix}-12D\\32\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{5\times 10-\left(-11\left(-2\right)\right)}&-\frac{-11}{5\times 10-\left(-11\left(-2\right)\right)}\\-\frac{-2}{5\times 10-\left(-11\left(-2\right)\right)}&\frac{5}{5\times 10-\left(-11\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}-12D\\32\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{14}&\frac{11}{28}\\\frac{1}{14}&\frac{5}{28}\end{matrix}\right)\left(\begin{matrix}-12D\\32\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{14}\left(-12D\right)+\frac{11}{28}\times 32\\\frac{1}{14}\left(-12D\right)+\frac{5}{28}\times 32\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{88-30D}{7}\\\frac{40-6D}{7}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{88-30D}{7},y=\frac{40-6D}{7}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2\left(x-y\right)-3\left(x+y\right)=-12D$$

Multiply 4 and 2 to get 8.

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

Use the distributive property to multiply 8 by x-y.

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

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

$$8x-8y-3x-3y=-12D$$

Combine 8x and -3x to get 5x.

$$5x-8y-3y=-12D$$

Combine -8y and -3y to get -11y.

$$5x-11y=-12D$$

Consider the second equation. Use the distributive property to multiply 6 by x+y.

$$6x+6y-4\left(2x-y\right)=16\times 2$$

Use the distributive property to multiply -4 by 2x-y.

$$6x+6y-8x+4y=16\times 2$$

Combine 6x and -8x to get -2x.

$$-2x+6y+4y=16\times 2$$

Combine 6y and 4y to get 10y.

$$-2x+10y=16\times 2$$

Multiply 16 and 2 to get 32.

$$-2x+10y=32$$

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.

$$5x-11y=-12D,-2x+10y=32$$

To make 5x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 5.

$$-2\times 5x-2\left(-11\right)y=-2\left(-12D\right),5\left(-2\right)x+5\times 10y=5\times 32$$

Simplify.

$$-10x+22y=24D,-10x+50y=160$$

Subtract -10x+50y=160 from -10x+22y=24D by subtracting like terms on each side of the equal sign.

$$-10x+10x+22y-50y=24D-160$$

Add -10x to 10x. Terms -10x and 10x cancel out, leaving an equation with only one variable that can be solved.

$$22y-50y=24D-160$$

Add 22y to -50y.

$$-28y=24D-160$$

Divide both sides by -28.

$$y=\frac{40-6D}{7}$$

Substitute \frac{40-6D}{7} for y in -2x+10y=32. Because the resulting equation contains only one variable, you can solve for x directly.

$$-2x+10\times \frac{40-6D}{7}=32$$

Multiply 10 times \frac{40-6D}{7}.

$$-2x+\frac{400-60D}{7}=32$$

Subtract \frac{400-60D}{7} from both sides of the equation.

$$-2x=\frac{60D-176}{7}$$

Divide both sides by -2.

$$x=\frac{88-30D}{7}$$

The system is now solved.

$$x=\frac{88-30D}{7},y=\frac{40-6D}{7}$$