Solve {l}{x:y=2}{x+y=10}

$$\left. \begin{array} { l } { x : y = 2 } \\ { x + y = 10 } \end{array} \right.$$

Solve for x, y

$$x = \frac{20}{3} = 6\frac{2}{3} \approx 6.666666667$$

$$y = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

Steps Using Substitution

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$x=2y$$

Substitute 2y for x in the other equation, x+y=10.

$$2y+y=10$$

Add 2y to y.

$$3y=10$$

Divide both sides by 3.

$$y=\frac{10}{3}$$

Substitute \frac{10}{3} for y in x=2y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=2\times \frac{10}{3}$$

Multiply 2 times \frac{10}{3}.

$$x=\frac{20}{3}$$

The system is now solved.

$$x=\frac{20}{3},y=\frac{10}{3}$$

Steps Using Matrices

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$x=2y$$

Subtract 2y from both sides.

$$x-2y=0$$

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

$$x-2y=0,x+y=10$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\10\end{matrix}\right)$$

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

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

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

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

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

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

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-2\right)}&-\frac{-2}{1-\left(-2\right)}\\-\frac{1}{1-\left(-2\right)}&\frac{1}{1-\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}0\\10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}&\frac{2}{3}\\-\frac{1}{3}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}0\\10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\times 10\\\frac{1}{3}\times 10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{3}\\\frac{10}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{20}{3},y=\frac{10}{3}$$

Steps Using Elimination

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$x=2y$$

Subtract 2y from both sides.

$$x-2y=0$$

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.

$$x-2y=0,x+y=10$$

Subtract x+y=10 from x-2y=0 by subtracting like terms on each side of the equal sign.

$$x-x-2y-y=-10$$

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

$$-2y-y=-10$$

Add -2y to -y.

$$-3y=-10$$

Divide both sides by -3.

$$y=\frac{10}{3}$$

Substitute \frac{10}{3} for y in x+y=10. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+\frac{10}{3}=10$$

Subtract \frac{10}{3} from both sides of the equation.

$$x=\frac{20}{3}$$

The system is now solved.

$$x=\frac{20}{3},y=\frac{10}{3}$$