Solve {l}{2(x+y/6)+2(x-y/3)=14/28}{x=y-x/3=frac{10}{12}}

$$\left. \begin{array} { l } { 2 ( \frac { x + y } { 6 } ) + 2 ( \frac { x - y } { 3 } ) = \frac { 14 } { 28 } } \\ { x = \frac { y - x } { 3 } = \frac { 10 } { 12 } } \end{array} \right.$$

Solve for x, y

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$y=-6$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 84, the least common multiple of 6,3,28.

$$28\left(x+y\right)+56\left(x-y\right)=3\times 14$$

Use the distributive property to multiply 28 by x+y.

$$28x+28y+56\left(x-y\right)=3\times 14$$

Use the distributive property to multiply 56 by x-y.

$$28x+28y+56x-56y=3\times 14$$

Combine 28x and 56x to get 84x.

$$84x+28y-56y=3\times 14$$

Combine 28y and -56y to get -28y.

$$84x-28y=3\times 14$$

Multiply 3 and 14 to get 42.

$$84x-28y=42$$

Consider the second equation. Divide each term of y-x by 3 to get \frac{1}{3}y-\frac{1}{3}x.

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

Subtract \frac{1}{3}y from both sides.

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

Add \frac{1}{3}x to both sides.

$$x-\frac{1}{3}y+\frac{1}{3}x=0$$

Combine x and \frac{1}{3}x to get \frac{4}{3}x.

$$\frac{4}{3}x-\frac{1}{3}y=0$$

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.

$$84x-28y=42,\frac{4}{3}x-\frac{1}{3}y=0$$

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

$$84x-28y=42$$

Add 28y to both sides of the equation.

$$84x=28y+42$$

Divide both sides by 84.

$$x=\frac{1}{84}\left(28y+42\right)$$

Multiply \frac{1}{84} times 28y+42.

$$x=\frac{1}{3}y+\frac{1}{2}$$

Substitute \frac{y}{3}+\frac{1}{2} for x in the other equation, \frac{4}{3}x-\frac{1}{3}y=0.

$$\frac{4}{3}\left(\frac{1}{3}y+\frac{1}{2}\right)-\frac{1}{3}y=0$$

Multiply \frac{4}{3} times \frac{y}{3}+\frac{1}{2}.

$$\frac{4}{9}y+\frac{2}{3}-\frac{1}{3}y=0$$

Add \frac{4y}{9} to -\frac{y}{3}.

$$\frac{1}{9}y+\frac{2}{3}=0$$

Subtract \frac{2}{3} from both sides of the equation.

$$\frac{1}{9}y=-\frac{2}{3}$$

Multiply both sides by 9.

$$y=-6$$

Substitute -6 for y in x=\frac{1}{3}y+\frac{1}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{3}\left(-6\right)+\frac{1}{2}$$

Multiply \frac{1}{3} times -6.

$$x=-2+\frac{1}{2}$$

Add \frac{1}{2} to -2.

$$x=-\frac{3}{2}$$

The system is now solved.

$$x=-\frac{3}{2},y=-6$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 84, the least common multiple of 6,3,28.

$$28\left(x+y\right)+56\left(x-y\right)=3\times 14$$

Use the distributive property to multiply 28 by x+y.

$$28x+28y+56\left(x-y\right)=3\times 14$$

Use the distributive property to multiply 56 by x-y.

$$28x+28y+56x-56y=3\times 14$$

Combine 28x and 56x to get 84x.

$$84x+28y-56y=3\times 14$$

Combine 28y and -56y to get -28y.

$$84x-28y=3\times 14$$

Multiply 3 and 14 to get 42.

$$84x-28y=42$$

Consider the second equation. Divide each term of y-x by 3 to get \frac{1}{3}y-\frac{1}{3}x.

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

Subtract \frac{1}{3}y from both sides.

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

Add \frac{1}{3}x to both sides.

$$x-\frac{1}{3}y+\frac{1}{3}x=0$$

Combine x and \frac{1}{3}x to get \frac{4}{3}x.

$$\frac{4}{3}x-\frac{1}{3}y=0$$

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

$$84x-28y=42,\frac{4}{3}x-\frac{1}{3}y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}42\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right).

$$inverse(\left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}42\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}42\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}84&-28\\\frac{4}{3}&-\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}42\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{1}{3}}{84\left(-\frac{1}{3}\right)-\left(-28\times \frac{4}{3}\right)}&-\frac{-28}{84\left(-\frac{1}{3}\right)-\left(-28\times \frac{4}{3}\right)}\\-\frac{\frac{4}{3}}{84\left(-\frac{1}{3}\right)-\left(-28\times \frac{4}{3}\right)}&\frac{84}{84\left(-\frac{1}{3}\right)-\left(-28\times \frac{4}{3}\right)}\end{matrix}\right)\left(\begin{matrix}42\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{28}&3\\-\frac{1}{7}&9\end{matrix}\right)\left(\begin{matrix}42\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{28}\times 42\\-\frac{1}{7}\times 42\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=-\frac{3}{2},y=-6$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 84, the least common multiple of 6,3,28.

$$28\left(x+y\right)+56\left(x-y\right)=3\times 14$$

Use the distributive property to multiply 28 by x+y.

$$28x+28y+56\left(x-y\right)=3\times 14$$

Use the distributive property to multiply 56 by x-y.

$$28x+28y+56x-56y=3\times 14$$

Combine 28x and 56x to get 84x.

$$84x+28y-56y=3\times 14$$

Combine 28y and -56y to get -28y.

$$84x-28y=3\times 14$$

Multiply 3 and 14 to get 42.

$$84x-28y=42$$

Consider the second equation. Divide each term of y-x by 3 to get \frac{1}{3}y-\frac{1}{3}x.

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

Subtract \frac{1}{3}y from both sides.

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

Add \frac{1}{3}x to both sides.

$$x-\frac{1}{3}y+\frac{1}{3}x=0$$

Combine x and \frac{1}{3}x to get \frac{4}{3}x.

$$\frac{4}{3}x-\frac{1}{3}y=0$$

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.

$$84x-28y=42,\frac{4}{3}x-\frac{1}{3}y=0$$

To make 84x and \frac{4x}{3} equal, multiply all terms on each side of the first equation by \frac{4}{3} and all terms on each side of the second by 84.

$$\frac{4}{3}\times 84x+\frac{4}{3}\left(-28\right)y=\frac{4}{3}\times 42,84\times \frac{4}{3}x+84\left(-\frac{1}{3}\right)y=0$$

Simplify.

$$112x-\frac{112}{3}y=56,112x-28y=0$$

Subtract 112x-28y=0 from 112x-\frac{112}{3}y=56 by subtracting like terms on each side of the equal sign.

$$112x-112x-\frac{112}{3}y+28y=56$$

Add 112x to -112x. Terms 112x and -112x cancel out, leaving an equation with only one variable that can be solved.

$$-\frac{112}{3}y+28y=56$$

Add -\frac{112y}{3} to 28y.

$$-\frac{28}{3}y=56$$

Divide both sides of the equation by -\frac{28}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-6$$

Substitute -6 for y in \frac{4}{3}x-\frac{1}{3}y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{4}{3}x-\frac{1}{3}\left(-6\right)=0$$

Multiply -\frac{1}{3} times -6.

$$\frac{4}{3}x+2=0$$

Subtract 2 from both sides of the equation.

$$\frac{4}{3}x=-2$$

Divide both sides of the equation by \frac{4}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=-\frac{3}{2}$$

The system is now solved.

$$x=-\frac{3}{2},y=-6$$