Solve {l}{14x+77y=1052}{22x+y=2054}

$$\left. \begin{array} { l } { 14 x + 77 y = 1052 } \\ { 22 x + y = 2054 } \end{array} \right.$$

Solve for x, y

$$x = \frac{78553}{840} = 93\frac{433}{840} \approx 93.51547619$$

$$y = -\frac{1403}{420} = -3\frac{143}{420} \approx -3.34047619$$

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.

$$14x+77y=1052,22x+y=2054$$

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

$$14x+77y=1052$$

Subtract 77y from both sides of the equation.

$$14x=-77y+1052$$

Divide both sides by 14.

$$x=\frac{1}{14}\left(-77y+1052\right)$$

Multiply \frac{1}{14} times -77y+1052.

$$x=-\frac{11}{2}y+\frac{526}{7}$$

Substitute -\frac{11y}{2}+\frac{526}{7} for x in the other equation, 22x+y=2054.

$$22\left(-\frac{11}{2}y+\frac{526}{7}\right)+y=2054$$

Multiply 22 times -\frac{11y}{2}+\frac{526}{7}.

$$-121y+\frac{11572}{7}+y=2054$$

Add -121y to y.

$$-120y+\frac{11572}{7}=2054$$

Subtract \frac{11572}{7} from both sides of the equation.

$$-120y=\frac{2806}{7}$$

Divide both sides by -120.

$$y=-\frac{1403}{420}$$

Substitute -\frac{1403}{420} for y in x=-\frac{11}{2}y+\frac{526}{7}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{11}{2}\left(-\frac{1403}{420}\right)+\frac{526}{7}$$

Multiply -\frac{11}{2} times -\frac{1403}{420} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{15433}{840}+\frac{526}{7}$$

Add \frac{526}{7} to \frac{15433}{840} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{78553}{840}$$

The system is now solved.

$$x=\frac{78553}{840},y=-\frac{1403}{420}$$

Steps Using Matrices

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

$$14x+77y=1052,22x+y=2054$$

Write the equations in matrix form.

$$\left(\begin{matrix}14&77\\22&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1052\\2054\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}14&77\\22&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}14&77\\22&1\end{matrix}\right))\left(\begin{matrix}14&77\\22&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&77\\22&1\end{matrix}\right))\left(\begin{matrix}1052\\2054\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&77\\22&1\end{matrix}\right))\left(\begin{matrix}1052\\2054\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&77\\22&1\end{matrix}\right))\left(\begin{matrix}1052\\2054\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{14-77\times 22}&-\frac{77}{14-77\times 22}\\-\frac{22}{14-77\times 22}&\frac{14}{14-77\times 22}\end{matrix}\right)\left(\begin{matrix}1052\\2054\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{1680}&\frac{11}{240}\\\frac{11}{840}&-\frac{1}{120}\end{matrix}\right)\left(\begin{matrix}1052\\2054\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{1680}\times 1052+\frac{11}{240}\times 2054\\\frac{11}{840}\times 1052-\frac{1}{120}\times 2054\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{78553}{840}\\-\frac{1403}{420}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{78553}{840},y=-\frac{1403}{420}$$

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.

$$14x+77y=1052,22x+y=2054$$

To make 14x and 22x equal, multiply all terms on each side of the first equation by 22 and all terms on each side of the second by 14.

$$22\times 14x+22\times 77y=22\times 1052,14\times 22x+14y=14\times 2054$$

Simplify.

$$308x+1694y=23144,308x+14y=28756$$

Subtract 308x+14y=28756 from 308x+1694y=23144 by subtracting like terms on each side of the equal sign.

$$308x-308x+1694y-14y=23144-28756$$

Add 308x to -308x. Terms 308x and -308x cancel out, leaving an equation with only one variable that can be solved.

$$1694y-14y=23144-28756$$

Add 1694y to -14y.

$$1680y=23144-28756$$

Add 23144 to -28756.

$$1680y=-5612$$

Divide both sides by 1680.

$$y=-\frac{1403}{420}$$

Substitute -\frac{1403}{420} for y in 22x+y=2054. Because the resulting equation contains only one variable, you can solve for x directly.

$$22x-\frac{1403}{420}=2054$$

Add \frac{1403}{420} to both sides of the equation.

$$22x=\frac{864083}{420}$$

Divide both sides by 22.

$$x=\frac{78553}{840}$$

The system is now solved.

$$x=\frac{78553}{840},y=-\frac{1403}{420}$$