Solve {l}{32x+50y=14600000}{10x+40y=7000000}

$$\left. \begin{array} { l } { 32 x + 50 y = 14600000 } \\ { 10 x + 40 y = 7000000 } \end{array} \right.$$

Solve for x, y

$$x=300000$$

$$y=100000$$

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.

$$32x+50y=14600000,10x+40y=7000000$$

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

$$32x+50y=14600000$$

Subtract 50y from both sides of the equation.

$$32x=-50y+14600000$$

Divide both sides by 32.

$$x=\frac{1}{32}\left(-50y+14600000\right)$$

Multiply \frac{1}{32} times -50y+14600000.

$$x=-\frac{25}{16}y+456250$$

Substitute -\frac{25y}{16}+456250 for x in the other equation, 10x+40y=7000000.

$$10\left(-\frac{25}{16}y+456250\right)+40y=7000000$$

Multiply 10 times -\frac{25y}{16}+456250.

$$-\frac{125}{8}y+4562500+40y=7000000$$

Add -\frac{125y}{8} to 40y.

$$\frac{195}{8}y+4562500=7000000$$

Subtract 4562500 from both sides of the equation.

$$\frac{195}{8}y=2437500$$

Divide both sides of the equation by \frac{195}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=100000$$

Substitute 100000 for y in x=-\frac{25}{16}y+456250. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{25}{16}\times 100000+456250$$

Multiply -\frac{25}{16} times 100000.

$$x=-156250+456250$$

Add 456250 to -156250.

$$x=300000$$

The system is now solved.

$$x=300000,y=100000$$

Steps Using Matrices

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

$$32x+50y=14600000,10x+40y=7000000$$

Write the equations in matrix form.

$$\left(\begin{matrix}32&50\\10&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14600000\\7000000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}32&50\\10&40\end{matrix}\right).

$$inverse(\left(\begin{matrix}32&50\\10&40\end{matrix}\right))\left(\begin{matrix}32&50\\10&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}32&50\\10&40\end{matrix}\right))\left(\begin{matrix}14600000\\7000000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}32&50\\10&40\end{matrix}\right))\left(\begin{matrix}14600000\\7000000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}32&50\\10&40\end{matrix}\right))\left(\begin{matrix}14600000\\7000000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40}{32\times 40-50\times 10}&-\frac{50}{32\times 40-50\times 10}\\-\frac{10}{32\times 40-50\times 10}&\frac{32}{32\times 40-50\times 10}\end{matrix}\right)\left(\begin{matrix}14600000\\7000000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{39}&-\frac{5}{78}\\-\frac{1}{78}&\frac{8}{195}\end{matrix}\right)\left(\begin{matrix}14600000\\7000000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{39}\times 14600000-\frac{5}{78}\times 7000000\\-\frac{1}{78}\times 14600000+\frac{8}{195}\times 7000000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300000\\100000\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=300000,y=100000$$

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.

$$32x+50y=14600000,10x+40y=7000000$$

To make 32x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 32.

$$10\times 32x+10\times 50y=10\times 14600000,32\times 10x+32\times 40y=32\times 7000000$$

Simplify.

$$320x+500y=146000000,320x+1280y=224000000$$

Subtract 320x+1280y=224000000 from 320x+500y=146000000 by subtracting like terms on each side of the equal sign.

$$320x-320x+500y-1280y=146000000-224000000$$

Add 320x to -320x. Terms 320x and -320x cancel out, leaving an equation with only one variable that can be solved.

$$500y-1280y=146000000-224000000$$

Add 500y to -1280y.

$$-780y=146000000-224000000$$

Add 146000000 to -224000000.

$$-780y=-78000000$$

Divide both sides by -780.

$$y=100000$$

Substitute 100000 for y in 10x+40y=7000000. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+40\times 100000=7000000$$

Multiply 40 times 100000.

$$10x+4000000=7000000$$

Subtract 4000000 from both sides of the equation.

$$10x=3000000$$

Divide both sides by 10.

$$x=300000$$

The system is now solved.

$$x=300000,y=100000$$