Solve {r}{4x+5y=153000}{-4x+4y=140000}

$$\left. \begin{array} { r } { 4 x + 5 y = 153000 } \\ { - 4 x + 4 y = 140000 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{22000}{9} = -2444\frac{4}{9} \approx -2444.444444444$$

$$y = \frac{293000}{9} = 32555\frac{5}{9} \approx 32555.555555556$$

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.

$$4x+5y=153000,-4x+4y=140000$$

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

$$4x+5y=153000$$

Subtract 5y from both sides of the equation.

$$4x=-5y+153000$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(-5y+153000\right)$$

Multiply \frac{1}{4} times -5y+153000.

$$x=-\frac{5}{4}y+38250$$

Substitute -\frac{5y}{4}+38250 for x in the other equation, -4x+4y=140000.

$$-4\left(-\frac{5}{4}y+38250\right)+4y=140000$$

Multiply -4 times -\frac{5y}{4}+38250.

$$5y-153000+4y=140000$$

Add 5y to 4y.

$$9y-153000=140000$$

Add 153000 to both sides of the equation.

$$9y=293000$$

Divide both sides by 9.

$$y=\frac{293000}{9}$$

Substitute \frac{293000}{9} for y in x=-\frac{5}{4}y+38250. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{4}\times \frac{293000}{9}+38250$$

Multiply -\frac{5}{4} times \frac{293000}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{366250}{9}+38250$$

Add 38250 to -\frac{366250}{9}.

$$x=-\frac{22000}{9}$$

The system is now solved.

$$x=-\frac{22000}{9},y=\frac{293000}{9}$$

Steps Using Matrices

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

$$4x+5y=153000,-4x+4y=140000$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&5\\-4&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}153000\\140000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&5\\-4&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&5\\-4&4\end{matrix}\right))\left(\begin{matrix}4&5\\-4&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&5\\-4&4\end{matrix}\right))\left(\begin{matrix}153000\\140000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&5\\-4&4\end{matrix}\right))\left(\begin{matrix}153000\\140000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&5\\-4&4\end{matrix}\right))\left(\begin{matrix}153000\\140000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4\times 4-5\left(-4\right)}&-\frac{5}{4\times 4-5\left(-4\right)}\\-\frac{-4}{4\times 4-5\left(-4\right)}&\frac{4}{4\times 4-5\left(-4\right)}\end{matrix}\right)\left(\begin{matrix}153000\\140000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}&-\frac{5}{36}\\\frac{1}{9}&\frac{1}{9}\end{matrix}\right)\left(\begin{matrix}153000\\140000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}\times 153000-\frac{5}{36}\times 140000\\\frac{1}{9}\times 153000+\frac{1}{9}\times 140000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{22000}{9}\\\frac{293000}{9}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{22000}{9},y=\frac{293000}{9}$$

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.

$$4x+5y=153000,-4x+4y=140000$$

To make 4x and -4x equal, multiply all terms on each side of the first equation by -4 and all terms on each side of the second by 4.

$$-4\times 4x-4\times 5y=-4\times 153000,4\left(-4\right)x+4\times 4y=4\times 140000$$

Simplify.

$$-16x-20y=-612000,-16x+16y=560000$$

Subtract -16x+16y=560000 from -16x-20y=-612000 by subtracting like terms on each side of the equal sign.

$$-16x+16x-20y-16y=-612000-560000$$

Add -16x to 16x. Terms -16x and 16x cancel out, leaving an equation with only one variable that can be solved.

$$-20y-16y=-612000-560000$$

Add -20y to -16y.

$$-36y=-612000-560000$$

Add -612000 to -560000.

$$-36y=-1172000$$

Divide both sides by -36.

$$y=\frac{293000}{9}$$

Substitute \frac{293000}{9} for y in -4x+4y=140000. Because the resulting equation contains only one variable, you can solve for x directly.

$$-4x+4\times \frac{293000}{9}=140000$$

Multiply 4 times \frac{293000}{9}.

$$-4x+\frac{1172000}{9}=140000$$

Subtract \frac{1172000}{9} from both sides of the equation.

$$-4x=\frac{88000}{9}$$

Divide both sides by -4.

$$x=-\frac{22000}{9}$$

The system is now solved.

$$x=-\frac{22000}{9},y=\frac{293000}{9}$$