Solve {c}{5000+x=2400y}{6y=5x}

$$\left. \begin{array} { c } { 5000 + x = 2400 y } \\ { 6 y = 5 x } \end{array} \right.$$

Solve for x, y

$$x = \frac{5000}{1999} = 2\frac{1002}{1999} \approx 2.501250625$$

$$y = \frac{12500}{5997} = 2\frac{506}{5997} \approx 2.084375521$$

Steps Using Substitution

Consider the first equation. Subtract 2400y from both sides.

$$5000+x-2400y=0$$

Subtract 5000 from both sides. Anything subtracted from zero gives its negation.

$$x-2400y=-5000$$

Consider the second equation. Subtract 5x from both sides.

$$6y-5x=0$$

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.

$$x-2400y=-5000,-5x+6y=0$$

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

$$x-2400y=-5000$$

Add 2400y to both sides of the equation.

$$x=2400y-5000$$

Substitute 2400y-5000 for x in the other equation, -5x+6y=0.

$$-5\left(2400y-5000\right)+6y=0$$

Multiply -5 times 2400y-5000.

$$-12000y+25000+6y=0$$

Add -12000y to 6y.

$$-11994y+25000=0$$

Subtract 25000 from both sides of the equation.

$$-11994y=-25000$$

Divide both sides by -11994.

$$y=\frac{12500}{5997}$$

Substitute \frac{12500}{5997} for y in x=2400y-5000. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=2400\times \frac{12500}{5997}-5000$$

Multiply 2400 times \frac{12500}{5997}.

$$x=\frac{10000000}{1999}-5000$$

Add -5000 to \frac{10000000}{1999}.

$$x=\frac{5000}{1999}$$

The system is now solved.

$$x=\frac{5000}{1999},y=\frac{12500}{5997}$$

Steps Using Matrices

Consider the first equation. Subtract 2400y from both sides.

$$5000+x-2400y=0$$

Subtract 5000 from both sides. Anything subtracted from zero gives its negation.

$$x-2400y=-5000$$

Consider the second equation. Subtract 5x from both sides.

$$6y-5x=0$$

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

$$x-2400y=-5000,-5x+6y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5000\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right))\left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right))\left(\begin{matrix}-5000\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right))\left(\begin{matrix}-5000\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2400\\-5&6\end{matrix}\right))\left(\begin{matrix}-5000\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{6-\left(-2400\left(-5\right)\right)}&-\frac{-2400}{6-\left(-2400\left(-5\right)\right)}\\-\frac{-5}{6-\left(-2400\left(-5\right)\right)}&\frac{1}{6-\left(-2400\left(-5\right)\right)}\end{matrix}\right)\left(\begin{matrix}-5000\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{1999}&-\frac{400}{1999}\\-\frac{5}{11994}&-\frac{1}{11994}\end{matrix}\right)\left(\begin{matrix}-5000\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{1999}\left(-5000\right)\\-\frac{5}{11994}\left(-5000\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5000}{1999}\\\frac{12500}{5997}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{5000}{1999},y=\frac{12500}{5997}$$

Steps Using Elimination

Consider the first equation. Subtract 2400y from both sides.

$$5000+x-2400y=0$$

Subtract 5000 from both sides. Anything subtracted from zero gives its negation.

$$x-2400y=-5000$$

Consider the second equation. Subtract 5x from both sides.

$$6y-5x=0$$

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.

$$x-2400y=-5000,-5x+6y=0$$

To make x and -5x equal, multiply all terms on each side of the first equation by -5 and all terms on each side of the second by 1.

$$-5x-5\left(-2400\right)y=-5\left(-5000\right),-5x+6y=0$$

Simplify.

$$-5x+12000y=25000,-5x+6y=0$$

Subtract -5x+6y=0 from -5x+12000y=25000 by subtracting like terms on each side of the equal sign.

$$-5x+5x+12000y-6y=25000$$

Add -5x to 5x. Terms -5x and 5x cancel out, leaving an equation with only one variable that can be solved.

$$12000y-6y=25000$$

Add 12000y to -6y.

$$11994y=25000$$

Divide both sides by 11994.

$$y=\frac{12500}{5997}$$

Substitute \frac{12500}{5997} for y in -5x+6y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-5x+6\times \frac{12500}{5997}=0$$

Multiply 6 times \frac{12500}{5997}.

$$-5x+\frac{25000}{1999}=0$$

Subtract \frac{25000}{1999} from both sides of the equation.

$$-5x=-\frac{25000}{1999}$$

Divide both sides by -5.

$$x=\frac{5000}{1999}$$

The system is now solved.

$$x=\frac{5000}{1999},y=\frac{12500}{5997}$$