Solve {l}{54x+30y=900}{32x+62y=1250}

$$\left. \begin{array} { l } { 54 x + 30 y = 900 } \\ { 32 x + 62 y = 1250 } \end{array} \right.$$

Solve for x, y

$$x = \frac{1525}{199} = 7\frac{132}{199} \approx 7.663316583$$

$$y = \frac{3225}{199} = 16\frac{41}{199} \approx 16.206030151$$

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.

$$54x+30y=900,32x+62y=1250$$

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

$$54x+30y=900$$

Subtract 30y from both sides of the equation.

$$54x=-30y+900$$

Divide both sides by 54.

$$x=\frac{1}{54}\left(-30y+900\right)$$

Multiply \frac{1}{54} times -30y+900.

$$x=-\frac{5}{9}y+\frac{50}{3}$$

Substitute -\frac{5y}{9}+\frac{50}{3} for x in the other equation, 32x+62y=1250.

$$32\left(-\frac{5}{9}y+\frac{50}{3}\right)+62y=1250$$

Multiply 32 times -\frac{5y}{9}+\frac{50}{3}.

$$-\frac{160}{9}y+\frac{1600}{3}+62y=1250$$

Add -\frac{160y}{9} to 62y.

$$\frac{398}{9}y+\frac{1600}{3}=1250$$

Subtract \frac{1600}{3} from both sides of the equation.

$$\frac{398}{9}y=\frac{2150}{3}$$

Divide both sides of the equation by \frac{398}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{3225}{199}$$

Substitute \frac{3225}{199} for y in x=-\frac{5}{9}y+\frac{50}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{9}\times \frac{3225}{199}+\frac{50}{3}$$

Multiply -\frac{5}{9} times \frac{3225}{199} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{5375}{597}+\frac{50}{3}$$

Add \frac{50}{3} to -\frac{5375}{597} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1525}{199}$$

The system is now solved.

$$x=\frac{1525}{199},y=\frac{3225}{199}$$

Steps Using Matrices

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

$$54x+30y=900,32x+62y=1250$$

Write the equations in matrix form.

$$\left(\begin{matrix}54&30\\32&62\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}900\\1250\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}54&30\\32&62\end{matrix}\right).

$$inverse(\left(\begin{matrix}54&30\\32&62\end{matrix}\right))\left(\begin{matrix}54&30\\32&62\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}54&30\\32&62\end{matrix}\right))\left(\begin{matrix}900\\1250\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}54&30\\32&62\end{matrix}\right))\left(\begin{matrix}900\\1250\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}54&30\\32&62\end{matrix}\right))\left(\begin{matrix}900\\1250\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{62}{54\times 62-30\times 32}&-\frac{30}{54\times 62-30\times 32}\\-\frac{32}{54\times 62-30\times 32}&\frac{54}{54\times 62-30\times 32}\end{matrix}\right)\left(\begin{matrix}900\\1250\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{31}{1194}&-\frac{5}{398}\\-\frac{8}{597}&\frac{9}{398}\end{matrix}\right)\left(\begin{matrix}900\\1250\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{31}{1194}\times 900-\frac{5}{398}\times 1250\\-\frac{8}{597}\times 900+\frac{9}{398}\times 1250\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1525}{199}\\\frac{3225}{199}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1525}{199},y=\frac{3225}{199}$$

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.

$$54x+30y=900,32x+62y=1250$$

To make 54x and 32x equal, multiply all terms on each side of the first equation by 32 and all terms on each side of the second by 54.

$$32\times 54x+32\times 30y=32\times 900,54\times 32x+54\times 62y=54\times 1250$$

Simplify.

$$1728x+960y=28800,1728x+3348y=67500$$

Subtract 1728x+3348y=67500 from 1728x+960y=28800 by subtracting like terms on each side of the equal sign.

$$1728x-1728x+960y-3348y=28800-67500$$

Add 1728x to -1728x. Terms 1728x and -1728x cancel out, leaving an equation with only one variable that can be solved.

$$960y-3348y=28800-67500$$

Add 960y to -3348y.

$$-2388y=28800-67500$$

Add 28800 to -67500.

$$-2388y=-38700$$

Divide both sides by -2388.

$$y=\frac{3225}{199}$$

Substitute \frac{3225}{199} for y in 32x+62y=1250. Because the resulting equation contains only one variable, you can solve for x directly.

$$32x+62\times \frac{3225}{199}=1250$$

Multiply 62 times \frac{3225}{199}.

$$32x+\frac{199950}{199}=1250$$

Subtract \frac{199950}{199} from both sides of the equation.

$$32x=\frac{48800}{199}$$

Divide both sides by 32.

$$x=\frac{1525}{199}$$

The system is now solved.

$$x=\frac{1525}{199},y=\frac{3225}{199}$$