Solve {l}{68x-72y=324}{32x+y=100}

$$\left\{ \begin{array} { l } { 68 x - 72 y = 324 } \\ { 32 x + y = 100 } \end{array} \right.$$

Solve for x, y

$$x = \frac{1881}{593} = 3\frac{102}{593} \approx 3.172006745$$

$$y = -\frac{892}{593} = -1\frac{299}{593} \approx -1.504215852$$

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.

$$68x-72y=324,32x+y=100$$

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

$$68x-72y=324$$

Add 72y to both sides of the equation.

$$68x=72y+324$$

Divide both sides by 68.

$$x=\frac{1}{68}\left(72y+324\right)$$

Multiply \frac{1}{68} times 72y+324.

$$x=\frac{18}{17}y+\frac{81}{17}$$

Substitute \frac{18y+81}{17} for x in the other equation, 32x+y=100.

$$32\left(\frac{18}{17}y+\frac{81}{17}\right)+y=100$$

Multiply 32 times \frac{18y+81}{17}.

$$\frac{576}{17}y+\frac{2592}{17}+y=100$$

Add \frac{576y}{17} to y.

$$\frac{593}{17}y+\frac{2592}{17}=100$$

Subtract \frac{2592}{17} from both sides of the equation.

$$\frac{593}{17}y=-\frac{892}{17}$$

Divide both sides of the equation by \frac{593}{17}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{892}{593}$$

Substitute -\frac{892}{593} for y in x=\frac{18}{17}y+\frac{81}{17}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{18}{17}\left(-\frac{892}{593}\right)+\frac{81}{17}$$

Multiply \frac{18}{17} times -\frac{892}{593} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{16056}{10081}+\frac{81}{17}$$

Add \frac{81}{17} to -\frac{16056}{10081} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1881}{593}$$

The system is now solved.

$$x=\frac{1881}{593},y=-\frac{892}{593}$$

Steps Using Matrices

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

$$68x-72y=324,32x+y=100$$

Write the equations in matrix form.

$$\left(\begin{matrix}68&-72\\32&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}324\\100\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}68&-72\\32&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}68&-72\\32&1\end{matrix}\right))\left(\begin{matrix}68&-72\\32&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}68&-72\\32&1\end{matrix}\right))\left(\begin{matrix}324\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}68&-72\\32&1\end{matrix}\right))\left(\begin{matrix}324\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}68&-72\\32&1\end{matrix}\right))\left(\begin{matrix}324\\100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{68-\left(-72\times 32\right)}&-\frac{-72}{68-\left(-72\times 32\right)}\\-\frac{32}{68-\left(-72\times 32\right)}&\frac{68}{68-\left(-72\times 32\right)}\end{matrix}\right)\left(\begin{matrix}324\\100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2372}&\frac{18}{593}\\-\frac{8}{593}&\frac{17}{593}\end{matrix}\right)\left(\begin{matrix}324\\100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2372}\times 324+\frac{18}{593}\times 100\\-\frac{8}{593}\times 324+\frac{17}{593}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1881}{593}\\-\frac{892}{593}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1881}{593},y=-\frac{892}{593}$$

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.

$$68x-72y=324,32x+y=100$$

To make 68x 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 68.

$$32\times 68x+32\left(-72\right)y=32\times 324,68\times 32x+68y=68\times 100$$

Simplify.

$$2176x-2304y=10368,2176x+68y=6800$$

Subtract 2176x+68y=6800 from 2176x-2304y=10368 by subtracting like terms on each side of the equal sign.

$$2176x-2176x-2304y-68y=10368-6800$$

Add 2176x to -2176x. Terms 2176x and -2176x cancel out, leaving an equation with only one variable that can be solved.

$$-2304y-68y=10368-6800$$

Add -2304y to -68y.

$$-2372y=10368-6800$$

Add 10368 to -6800.

$$-2372y=3568$$

Divide both sides by -2372.

$$y=-\frac{892}{593}$$

Substitute -\frac{892}{593} for y in 32x+y=100. Because the resulting equation contains only one variable, you can solve for x directly.

$$32x-\frac{892}{593}=100$$

Add \frac{892}{593} to both sides of the equation.

$$32x=\frac{60192}{593}$$

Divide both sides by 32.

$$x=\frac{1881}{593}$$

The system is now solved.

$$x=\frac{1881}{593},y=-\frac{892}{593}$$