Solve {c}{4x+160y=128}{x+2y0y=82}

$$\left. \begin{array} { c } { 4 x + 160 y = 128 } \\ { x + 2 y 0 y = 82 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{8\left(205-4y_{0}\right)}{y_{0}-20}$$

$$y=\frac{25}{y_{0}-20}$$

$$y_{0}\neq 20$$

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.

$$4x+160y=128,x+2y_{0}y=82$$

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

$$4x+160y=128$$

Subtract 160y from both sides of the equation.

$$4x=-160y+128$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(-160y+128\right)$$

Multiply \frac{1}{4} times -160y+128.

$$x=-40y+32$$

Substitute -40y+32 for x in the other equation, x+2y_{0}y=82.

$$-40y+32+2y_{0}y=82$$

Add -40y to 2y_{0}y.

$$\left(2y_{0}-40\right)y+32=82$$

Subtract 32 from both sides of the equation.

$$\left(2y_{0}-40\right)y=50$$

Divide both sides by -40+2y_{0}.

$$y=\frac{25}{y_{0}-20}$$

Substitute \frac{25}{-20+y_{0}} for y in x=-40y+32. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-40\times \frac{25}{y_{0}-20}+32$$

Multiply -40 times \frac{25}{-20+y_{0}}.

$$x=-\frac{1000}{y_{0}-20}+32$$

Add 32 to -\frac{1000}{-20+y_{0}}.

$$x=\frac{8\left(4y_{0}-205\right)}{y_{0}-20}$$

The system is now solved.

$$x=\frac{8\left(4y_{0}-205\right)}{y_{0}-20},y=\frac{25}{y_{0}-20}$$

Steps Using Matrices

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

$$4x+160y=128,x+2y_{0}y=82$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}128\\82\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right))\left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right))\left(\begin{matrix}128\\82\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right))\left(\begin{matrix}128\\82\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&160\\1&2y_{0}\end{matrix}\right))\left(\begin{matrix}128\\82\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2y_{0}}{4\times 2y_{0}-160}&-\frac{160}{4\times 2y_{0}-160}\\-\frac{1}{4\times 2y_{0}-160}&\frac{4}{4\times 2y_{0}-160}\end{matrix}\right)\left(\begin{matrix}128\\82\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{y_{0}}{4\left(y_{0}-20\right)}&-\frac{20}{y_{0}-20}\\-\frac{1}{8\left(y_{0}-20\right)}&\frac{1}{2\left(y_{0}-20\right)}\end{matrix}\right)\left(\begin{matrix}128\\82\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{y_{0}}{4\left(y_{0}-20\right)}\times 128+\left(-\frac{20}{y_{0}-20}\right)\times 82\\\left(-\frac{1}{8\left(y_{0}-20\right)}\right)\times 128+\frac{1}{2\left(y_{0}-20\right)}\times 82\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8\left(4y_{0}-205\right)}{y_{0}-20}\\\frac{25}{y_{0}-20}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{8\left(4y_{0}-205\right)}{y_{0}-20},y=\frac{25}{y_{0}-20}$$

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.

$$4x+160y=128,x+2y_{0}y=82$$

To make 4x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 4.

$$4x+160y=128,4x+4\times 2y_{0}y=4\times 82$$

Simplify.

$$4x+160y=128,4x+8y_{0}y=328$$

Subtract 4x+8y_{0}y=328 from 4x+160y=128 by subtracting like terms on each side of the equal sign.

$$4x-4x+160y+\left(-8y_{0}\right)y=128-328$$

Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.

$$160y+\left(-8y_{0}\right)y=128-328$$

Add 160y to -8y_{0}y.

$$\left(160-8y_{0}\right)y=128-328$$

Add 128 to -328.

$$\left(160-8y_{0}\right)y=-200$$

Divide both sides by 160-8y_{0}.

$$y=-\frac{25}{20-y_{0}}$$

Substitute -\frac{25}{20-y_{0}} for y in x+2y_{0}y=82. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+2y_{0}\left(-\frac{25}{20-y_{0}}\right)=82$$

Multiply 2y_{0} times -\frac{25}{20-y_{0}}.

$$x-\frac{50y_{0}}{20-y_{0}}=82$$

Add \frac{50y_{0}}{20-y_{0}} to both sides of the equation.

$$x=\frac{8\left(205-4y_{0}\right)}{20-y_{0}}$$

The system is now solved.

$$x=\frac{8\left(205-4y_{0}\right)}{20-y_{0}},y=-\frac{25}{20-y_{0}}$$