Solve {l}{x+2y=14}{x+2x/y-1=3/2}

$$\left\{ \begin{array} { l } { x + 2 y = 14 } \\ { \frac { x + 2 x } { y - 1 } = \frac { 3 } { 2 } } \end{array} \right.$$

Solve for x, y

$$x = \frac{12}{5} = 2\frac{2}{5} = 2.4$$

$$y = \frac{29}{5} = 5\frac{4}{5} = 5.8$$

Steps Using Substitution

Consider the second equation. Variable y cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(y-1\right), the least common multiple of y-1,2.

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

Combine x and 2x to get 3x.

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

Multiply 2 and 3 to get 6.

$$6x=3\left(y-1\right)$$

Use the distributive property to multiply 3 by y-1.

$$6x=3y-3$$

Subtract 3y from both sides.

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

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.

$$x+2y=14,6x-3y=-3$$

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

$$x+2y=14$$

Subtract 2y from both sides of the equation.

$$x=-2y+14$$

Substitute 14-2y for x in the other equation, 6x-3y=-3.

$$6\left(-2y+14\right)-3y=-3$$

Multiply 6 times 14-2y.

$$-12y+84-3y=-3$$

Add -12y to -3y.

$$-15y+84=-3$$

Subtract 84 from both sides of the equation.

$$-15y=-87$$

Divide both sides by -15.

$$y=\frac{29}{5}$$

Substitute \frac{29}{5} for y in x=-2y+14. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times \frac{29}{5}+14$$

Multiply -2 times \frac{29}{5}.

$$x=-\frac{58}{5}+14$$

Add 14 to -\frac{58}{5}.

$$x=\frac{12}{5}$$

The system is now solved.

$$x=\frac{12}{5},y=\frac{29}{5}$$

Steps Using Matrices

Consider the second equation. Variable y cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(y-1\right), the least common multiple of y-1,2.

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

Combine x and 2x to get 3x.

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

Multiply 2 and 3 to get 6.

$$6x=3\left(y-1\right)$$

Use the distributive property to multiply 3 by y-1.

$$6x=3y-3$$

Subtract 3y from both sides.

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

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

$$x+2y=14,6x-3y=-3$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&2\\6&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\-3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\6&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&2\\6&-3\end{matrix}\right))\left(\begin{matrix}1&2\\6&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\6&-3\end{matrix}\right))\left(\begin{matrix}14\\-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\6&-3\end{matrix}\right))\left(\begin{matrix}14\\-3\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\6&-3\end{matrix}\right))\left(\begin{matrix}14\\-3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-2\times 6}&-\frac{2}{-3-2\times 6}\\-\frac{6}{-3-2\times 6}&\frac{1}{-3-2\times 6}\end{matrix}\right)\left(\begin{matrix}14\\-3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&\frac{2}{15}\\\frac{2}{5}&-\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}14\\-3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 14+\frac{2}{15}\left(-3\right)\\\frac{2}{5}\times 14-\frac{1}{15}\left(-3\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{12}{5}\\\frac{29}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{12}{5},y=\frac{29}{5}$$

Steps Using Elimination

Consider the second equation. Variable y cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(y-1\right), the least common multiple of y-1,2.

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

Combine x and 2x to get 3x.

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

Multiply 2 and 3 to get 6.

$$6x=3\left(y-1\right)$$

Use the distributive property to multiply 3 by y-1.

$$6x=3y-3$$

Subtract 3y from both sides.

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

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.

$$x+2y=14,6x-3y=-3$$

To make x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 1.

$$6x+6\times 2y=6\times 14,6x-3y=-3$$

Simplify.

$$6x+12y=84,6x-3y=-3$$

Subtract 6x-3y=-3 from 6x+12y=84 by subtracting like terms on each side of the equal sign.

$$6x-6x+12y+3y=84+3$$

Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.

$$12y+3y=84+3$$

Add 12y to 3y.

$$15y=84+3$$

Add 84 to 3.

$$15y=87$$

Divide both sides by 15.

$$y=\frac{29}{5}$$

Substitute \frac{29}{5} for y in 6x-3y=-3. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x-3\times \frac{29}{5}=-3$$

Multiply -3 times \frac{29}{5}.

$$6x-\frac{87}{5}=-3$$

Add \frac{87}{5} to both sides of the equation.

$$6x=\frac{72}{5}$$

Divide both sides by 6.

$$x=\frac{12}{5}$$

The system is now solved.

$$x=\frac{12}{5},y=\frac{29}{5}$$