Solve {l}{15x-5y=4000}{9x-2y=4000}

$$\left. \begin{array} { l } { 15 x - 5 y = 4000 } \\ { 9 x - 2 y = 4000 } \end{array} \right.$$

Solve for x, y

$$x=800$$

$$y=1600$$

Steps Using Substitution

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.

$$15x-5y=4000,9x-2y=4000$$

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

$$15x-5y=4000$$

Add 5y to both sides of the equation.

$$15x=5y+4000$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(5y+4000\right)$$

Multiply \frac{1}{15} times 4000+5y.

$$x=\frac{1}{3}y+\frac{800}{3}$$

Substitute \frac{800+y}{3} for x in the other equation, 9x-2y=4000.

$$9\left(\frac{1}{3}y+\frac{800}{3}\right)-2y=4000$$

Multiply 9 times \frac{800+y}{3}.

$$3y+2400-2y=4000$$

Add 3y to -2y.

$$y+2400=4000$$

Subtract 2400 from both sides of the equation.

$$y=1600$$

Substitute 1600 for y in x=\frac{1}{3}y+\frac{800}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{3}\times 1600+\frac{800}{3}$$

Multiply \frac{1}{3} times 1600.

$$x=\frac{1600+800}{3}$$

Add \frac{800}{3} to \frac{1600}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=800$$

The system is now solved.

$$x=800,y=1600$$

Steps Using Matrices

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

$$15x-5y=4000,9x-2y=4000$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&-5\\9&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4000\\4000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&-5\\9&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&-5\\9&-2\end{matrix}\right))\left(\begin{matrix}15&-5\\9&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-5\\9&-2\end{matrix}\right))\left(\begin{matrix}4000\\4000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-5\\9&-2\end{matrix}\right))\left(\begin{matrix}4000\\4000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-5\\9&-2\end{matrix}\right))\left(\begin{matrix}4000\\4000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{15\left(-2\right)-\left(-5\times 9\right)}&-\frac{-5}{15\left(-2\right)-\left(-5\times 9\right)}\\-\frac{9}{15\left(-2\right)-\left(-5\times 9\right)}&\frac{15}{15\left(-2\right)-\left(-5\times 9\right)}\end{matrix}\right)\left(\begin{matrix}4000\\4000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{15}&\frac{1}{3}\\-\frac{3}{5}&1\end{matrix}\right)\left(\begin{matrix}4000\\4000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{15}\times 4000+\frac{1}{3}\times 4000\\-\frac{3}{5}\times 4000+4000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}800\\1600\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=800,y=1600$$

Steps Using Elimination

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.

$$15x-5y=4000,9x-2y=4000$$

To make 15x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by 15.

$$9\times 15x+9\left(-5\right)y=9\times 4000,15\times 9x+15\left(-2\right)y=15\times 4000$$

Simplify.

$$135x-45y=36000,135x-30y=60000$$

Subtract 135x-30y=60000 from 135x-45y=36000 by subtracting like terms on each side of the equal sign.

$$135x-135x-45y+30y=36000-60000$$

Add 135x to -135x. Terms 135x and -135x cancel out, leaving an equation with only one variable that can be solved.

$$-45y+30y=36000-60000$$

Add -45y to 30y.

$$-15y=36000-60000$$

Add 36000 to -60000.

$$-15y=-24000$$

Divide both sides by -15.

$$y=1600$$

Substitute 1600 for y in 9x-2y=4000. Because the resulting equation contains only one variable, you can solve for x directly.

$$9x-2\times 1600=4000$$

Multiply -2 times 1600.

$$9x-3200=4000$$

Add 3200 to both sides of the equation.

$$9x=7200$$

Divide both sides by 9.

$$x=800$$

The system is now solved.

$$x=800,y=1600$$