Solve {l}{41x+53y=135}{53x+41y=147}

$$\left. \begin{array} { l } { 41 x + 53 y = 135 } \\ { 53 x + 41 y = 147 } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=1$$

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.

$$41x+53y=135,53x+41y=147$$

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

$$41x+53y=135$$

Subtract 53y from both sides of the equation.

$$41x=-53y+135$$

Divide both sides by 41.

$$x=\frac{1}{41}\left(-53y+135\right)$$

Multiply \frac{1}{41} times -53y+135.

$$x=-\frac{53}{41}y+\frac{135}{41}$$

Substitute \frac{-53y+135}{41} for x in the other equation, 53x+41y=147.

$$53\left(-\frac{53}{41}y+\frac{135}{41}\right)+41y=147$$

Multiply 53 times \frac{-53y+135}{41}.

$$-\frac{2809}{41}y+\frac{7155}{41}+41y=147$$

Add -\frac{2809y}{41} to 41y.

$$-\frac{1128}{41}y+\frac{7155}{41}=147$$

Subtract \frac{7155}{41} from both sides of the equation.

$$-\frac{1128}{41}y=-\frac{1128}{41}$$

Divide both sides of the equation by -\frac{1128}{41}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=1$$

Substitute 1 for y in x=-\frac{53}{41}y+\frac{135}{41}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{-53+135}{41}$$

Add \frac{135}{41} to -\frac{53}{41} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=2$$

The system is now solved.

$$x=2,y=1$$

Steps Using Matrices

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

$$41x+53y=135,53x+41y=147$$

Write the equations in matrix form.

$$\left(\begin{matrix}41&53\\53&41\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}135\\147\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}41&53\\53&41\end{matrix}\right).

$$inverse(\left(\begin{matrix}41&53\\53&41\end{matrix}\right))\left(\begin{matrix}41&53\\53&41\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}41&53\\53&41\end{matrix}\right))\left(\begin{matrix}135\\147\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}41&53\\53&41\end{matrix}\right))\left(\begin{matrix}135\\147\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}41&53\\53&41\end{matrix}\right))\left(\begin{matrix}135\\147\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{41}{41\times 41-53\times 53}&-\frac{53}{41\times 41-53\times 53}\\-\frac{53}{41\times 41-53\times 53}&\frac{41}{41\times 41-53\times 53}\end{matrix}\right)\left(\begin{matrix}135\\147\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{41}{1128}&\frac{53}{1128}\\\frac{53}{1128}&-\frac{41}{1128}\end{matrix}\right)\left(\begin{matrix}135\\147\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{41}{1128}\times 135+\frac{53}{1128}\times 147\\\frac{53}{1128}\times 135-\frac{41}{1128}\times 147\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2,y=1$$

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.

$$41x+53y=135,53x+41y=147$$

To make 41x and 53x equal, multiply all terms on each side of the first equation by 53 and all terms on each side of the second by 41.

$$53\times 41x+53\times 53y=53\times 135,41\times 53x+41\times 41y=41\times 147$$

Simplify.

$$2173x+2809y=7155,2173x+1681y=6027$$

Subtract 2173x+1681y=6027 from 2173x+2809y=7155 by subtracting like terms on each side of the equal sign.

$$2173x-2173x+2809y-1681y=7155-6027$$

Add 2173x to -2173x. Terms 2173x and -2173x cancel out, leaving an equation with only one variable that can be solved.

$$2809y-1681y=7155-6027$$

Add 2809y to -1681y.

$$1128y=7155-6027$$

Add 7155 to -6027.

$$1128y=1128$$

Divide both sides by 1128.

$$y=1$$

Substitute 1 for y in 53x+41y=147. Because the resulting equation contains only one variable, you can solve for x directly.

$$53x+41=147$$

Subtract 41 from both sides of the equation.

$$53x=106$$

Divide both sides by 53.

$$x=2$$

The system is now solved.

$$x=2,y=1$$