Solve {l}{x+y=30}{300y=200x}

$$\left\{ \begin{array} { l } { x + y = 30 } \\ { 300 y = 200 x } \end{array} \right.$$

Solve for x, y

$$x=18$$

$$y=12$$

Steps Using Substitution

Consider the second equation. Subtract 200x from both sides.

$$300y-200x=0$$

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.

$$x+y=30,-200x+300y=0$$

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

$$x+y=30$$

Subtract y from both sides of the equation.

$$x=-y+30$$

Substitute -y+30 for x in the other equation, -200x+300y=0.

$$-200\left(-y+30\right)+300y=0$$

Multiply -200 times -y+30.

$$200y-6000+300y=0$$

Add 200y to 300y.

$$500y-6000=0$$

Add 6000 to both sides of the equation.

$$500y=6000$$

Divide both sides by 500.

$$y=12$$

Substitute 12 for y in x=-y+30. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-12+30$$

Add 30 to -12.

$$x=18$$

The system is now solved.

$$x=18,y=12$$

Steps Using Matrices

Consider the second equation. Subtract 200x from both sides.

$$300y-200x=0$$

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

$$x+y=30,-200x+300y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\-200&300\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}30\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-200&300\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\-200&300\end{matrix}\right))\left(\begin{matrix}1&1\\-200&300\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-200&300\end{matrix}\right))\left(\begin{matrix}30\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-200&300\end{matrix}\right))\left(\begin{matrix}30\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-200&300\end{matrix}\right))\left(\begin{matrix}30\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{300}{300-\left(-200\right)}&-\frac{1}{300-\left(-200\right)}\\-\frac{-200}{300-\left(-200\right)}&\frac{1}{300-\left(-200\right)}\end{matrix}\right)\left(\begin{matrix}30\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}&-\frac{1}{500}\\\frac{2}{5}&\frac{1}{500}\end{matrix}\right)\left(\begin{matrix}30\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}\times 30\\\frac{2}{5}\times 30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18\\12\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=18,y=12$$

Steps Using Elimination

Consider the second equation. Subtract 200x from both sides.

$$300y-200x=0$$

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.

$$x+y=30,-200x+300y=0$$

To make x and -200x equal, multiply all terms on each side of the first equation by -200 and all terms on each side of the second by 1.

$$-200x-200y=-200\times 30,-200x+300y=0$$

Simplify.

$$-200x-200y=-6000,-200x+300y=0$$

Subtract -200x+300y=0 from -200x-200y=-6000 by subtracting like terms on each side of the equal sign.

$$-200x+200x-200y-300y=-6000$$

Add -200x to 200x. Terms -200x and 200x cancel out, leaving an equation with only one variable that can be solved.

$$-200y-300y=-6000$$

Add -200y to -300y.

$$-500y=-6000$$

Divide both sides by -500.

$$y=12$$

Substitute 12 for y in -200x+300y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-200x+300\times 12=0$$

Multiply 300 times 12.

$$-200x+3600=0$$

Subtract 3600 from both sides of the equation.

$$-200x=-3600$$

Divide both sides by -200.

$$x=18$$

The system is now solved.

$$x=18,y=12$$