Solve {l}{380x+320y=380y+320x}{2300y+320x=2000-180}

$$\left\{ \begin{array} { l } { 380 x + 320 y = 380 y + 320 x } \\ { 2300 y + 320 x = 2000 - 180 } \end{array} \right.$$

Solve for x, y

$$x=\frac{91}{131}\approx 0.694656489$$

$$y=\frac{91}{131}\approx 0.694656489$$

Steps Using Substitution

Consider the first equation. Subtract 380y from both sides.

$$380x+320y-380y=320x$$

Combine 320y and -380y to get -60y.

$$380x-60y=320x$$

Subtract 320x from both sides.

$$380x-60y-320x=0$$

Combine 380x and -320x to get 60x.

$$60x-60y=0$$

Consider the second equation. Subtract 180 from 2000 to get 1820.

$$2300y+320x=1820$$

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.

$$60x-60y=0,320x+2300y=1820$$

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

$$60x-60y=0$$

Add 60y to both sides of the equation.

$$60x=60y$$

Divide both sides by 60.

$$x=\frac{1}{60}\times 60y$$

Multiply \frac{1}{60} times 60y.

$$x=y$$

Substitute y for x in the other equation, 320x+2300y=1820.

$$320y+2300y=1820$$

Add 320y to 2300y.

$$2620y=1820$$

Divide both sides by 2620.

$$y=\frac{91}{131}$$

Substitute \frac{91}{131} for y in x=y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{91}{131}$$

The system is now solved.

$$x=\frac{91}{131},y=\frac{91}{131}$$

Steps Using Matrices

Consider the first equation. Subtract 380y from both sides.

$$380x+320y-380y=320x$$

Combine 320y and -380y to get -60y.

$$380x-60y=320x$$

Subtract 320x from both sides.

$$380x-60y-320x=0$$

Combine 380x and -320x to get 60x.

$$60x-60y=0$$

Consider the second equation. Subtract 180 from 2000 to get 1820.

$$2300y+320x=1820$$

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

$$60x-60y=0,320x+2300y=1820$$

Write the equations in matrix form.

$$\left(\begin{matrix}60&-60\\320&2300\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\1820\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}60&-60\\320&2300\end{matrix}\right).

$$inverse(\left(\begin{matrix}60&-60\\320&2300\end{matrix}\right))\left(\begin{matrix}60&-60\\320&2300\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}60&-60\\320&2300\end{matrix}\right))\left(\begin{matrix}0\\1820\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}60&-60\\320&2300\end{matrix}\right))\left(\begin{matrix}0\\1820\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}60&-60\\320&2300\end{matrix}\right))\left(\begin{matrix}0\\1820\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2300}{60\times 2300-\left(-60\times 320\right)}&-\frac{-60}{60\times 2300-\left(-60\times 320\right)}\\-\frac{320}{60\times 2300-\left(-60\times 320\right)}&\frac{60}{60\times 2300-\left(-60\times 320\right)}\end{matrix}\right)\left(\begin{matrix}0\\1820\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{23}{1572}&\frac{1}{2620}\\-\frac{4}{1965}&\frac{1}{2620}\end{matrix}\right)\left(\begin{matrix}0\\1820\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2620}\times 1820\\\frac{1}{2620}\times 1820\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{91}{131}\\\frac{91}{131}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{91}{131},y=\frac{91}{131}$$

Steps Using Elimination

Consider the first equation. Subtract 380y from both sides.

$$380x+320y-380y=320x$$

Combine 320y and -380y to get -60y.

$$380x-60y=320x$$

Subtract 320x from both sides.

$$380x-60y-320x=0$$

Combine 380x and -320x to get 60x.

$$60x-60y=0$$

Consider the second equation. Subtract 180 from 2000 to get 1820.

$$2300y+320x=1820$$

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.

$$60x-60y=0,320x+2300y=1820$$

To make 60x and 320x equal, multiply all terms on each side of the first equation by 320 and all terms on each side of the second by 60.

$$320\times 60x+320\left(-60\right)y=0,60\times 320x+60\times 2300y=60\times 1820$$

Simplify.

$$19200x-19200y=0,19200x+138000y=109200$$

Subtract 19200x+138000y=109200 from 19200x-19200y=0 by subtracting like terms on each side of the equal sign.

$$19200x-19200x-19200y-138000y=-109200$$

Add 19200x to -19200x. Terms 19200x and -19200x cancel out, leaving an equation with only one variable that can be solved.

$$-19200y-138000y=-109200$$

Add -19200y to -138000y.

$$-157200y=-109200$$

Divide both sides by -157200.

$$y=\frac{91}{131}$$

Substitute \frac{91}{131} for y in 320x+2300y=1820. Because the resulting equation contains only one variable, you can solve for x directly.

$$320x+2300\times \frac{91}{131}=1820$$

Multiply 2300 times \frac{91}{131}.

$$320x+\frac{209300}{131}=1820$$

Subtract \frac{209300}{131} from both sides of the equation.

$$320x=\frac{29120}{131}$$

Divide both sides by 320.

$$x=\frac{91}{131}$$

The system is now solved.

$$x=\frac{91}{131},y=\frac{91}{131}$$