Solve {c}{x+y=1.0}{891x+286y=488}

$$\left. \begin{array} { c } { x + y = 1.0 } \\ { 891 x + 286 y = 488 } \end{array} \right.$$

Solve for x, y

$$x=\frac{202}{605}\approx 0.333884298$$

$$y=\frac{403}{605}\approx 0.666115702$$

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.

$$x+y=1,891x+286y=488$$

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

$$x+y=1$$

Subtract y from both sides of the equation.

$$x=-y+1$$

Substitute -y+1 for x in the other equation, 891x+286y=488.

$$891\left(-y+1\right)+286y=488$$

Multiply 891 times -y+1.

$$-891y+891+286y=488$$

Add -891y to 286y.

$$-605y+891=488$$

Subtract 891 from both sides of the equation.

$$-605y=-403$$

Divide both sides by -605.

$$y=\frac{403}{605}$$

Substitute \frac{403}{605} for y in x=-y+1. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{403}{605}+1$$

Add 1 to -\frac{403}{605}.

$$x=\frac{202}{605}$$

The system is now solved.

$$x=\frac{202}{605},y=\frac{403}{605}$$

Steps Using Matrices

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

$$x+y=1,891x+286y=488$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\891&286\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\488\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\891&286\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\891&286\end{matrix}\right))\left(\begin{matrix}1&1\\891&286\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\891&286\end{matrix}\right))\left(\begin{matrix}1\\488\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\891&286\end{matrix}\right))\left(\begin{matrix}1\\488\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\891&286\end{matrix}\right))\left(\begin{matrix}1\\488\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{286}{286-891}&-\frac{1}{286-891}\\-\frac{891}{286-891}&\frac{1}{286-891}\end{matrix}\right)\left(\begin{matrix}1\\488\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{26}{55}&\frac{1}{605}\\\frac{81}{55}&-\frac{1}{605}\end{matrix}\right)\left(\begin{matrix}1\\488\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{26}{55}+\frac{1}{605}\times 488\\\frac{81}{55}-\frac{1}{605}\times 488\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{202}{605}\\\frac{403}{605}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{202}{605},y=\frac{403}{605}$$

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.

$$x+y=1,891x+286y=488$$

To make x and 891x equal, multiply all terms on each side of the first equation by 891 and all terms on each side of the second by 1.

$$891x+891y=891,891x+286y=488$$

Subtract 891x+286y=488 from 891x+891y=891 by subtracting like terms on each side of the equal sign.

$$891x-891x+891y-286y=891-488$$

Add 891x to -891x. Terms 891x and -891x cancel out, leaving an equation with only one variable that can be solved.

$$891y-286y=891-488$$

Add 891y to -286y.

$$605y=891-488$$

Add 891 to -488.

$$605y=403$$

Divide both sides by 605.

$$y=\frac{403}{605}$$

Substitute \frac{403}{605} for y in 891x+286y=488. Because the resulting equation contains only one variable, you can solve for x directly.

$$891x+286\times \frac{403}{605}=488$$

Multiply 286 times \frac{403}{605}.

$$891x+\frac{10478}{55}=488$$

Subtract \frac{10478}{55} from both sides of the equation.

$$891x=\frac{16362}{55}$$

Divide both sides by 891.

$$x=\frac{202}{605}$$

The system is now solved.

$$x=\frac{202}{605},y=\frac{403}{605}$$