Solve {l}{x-y=21}{x-y/20=16}

$$\left. \begin{array} { l } { x - y = 21 } \\ { x - \frac { y } { 20 } = 16 } \end{array} \right.$$

Solve for x, y

$$x = \frac{299}{19} = 15\frac{14}{19} \approx 15.736842105$$

$$y = -\frac{100}{19} = -5\frac{5}{19} \approx -5.263157895$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 20.

$$20x-y=320$$

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.

$$x-y=21,20x-y=320$$

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

$$x-y=21$$

Add y to both sides of the equation.

$$x=y+21$$

Substitute y+21 for x in the other equation, 20x-y=320.

$$20\left(y+21\right)-y=320$$

Multiply 20 times y+21.

$$20y+420-y=320$$

Add 20y to -y.

$$19y+420=320$$

Subtract 420 from both sides of the equation.

$$19y=-100$$

Divide both sides by 19.

$$y=-\frac{100}{19}$$

Substitute -\frac{100}{19} for y in x=y+21. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{100}{19}+21$$

Add 21 to -\frac{100}{19}.

$$x=\frac{299}{19}$$

The system is now solved.

$$x=\frac{299}{19},y=-\frac{100}{19}$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 20.

$$20x-y=320$$

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

$$x-y=21,20x-y=320$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\20&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}21\\320\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\20&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\20&-1\end{matrix}\right))\left(\begin{matrix}1&-1\\20&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\20&-1\end{matrix}\right))\left(\begin{matrix}21\\320\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\20&-1\end{matrix}\right))\left(\begin{matrix}21\\320\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\20&-1\end{matrix}\right))\left(\begin{matrix}21\\320\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-1-\left(-20\right)}&-\frac{-1}{-1-\left(-20\right)}\\-\frac{20}{-1-\left(-20\right)}&\frac{1}{-1-\left(-20\right)}\end{matrix}\right)\left(\begin{matrix}21\\320\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{19}&\frac{1}{19}\\-\frac{20}{19}&\frac{1}{19}\end{matrix}\right)\left(\begin{matrix}21\\320\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{19}\times 21+\frac{1}{19}\times 320\\-\frac{20}{19}\times 21+\frac{1}{19}\times 320\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{299}{19}\\-\frac{100}{19}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{299}{19},y=-\frac{100}{19}$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 20.

$$20x-y=320$$

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.

$$x-y=21,20x-y=320$$

Subtract 20x-y=320 from x-y=21 by subtracting like terms on each side of the equal sign.

$$x-20x-y+y=21-320$$

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

$$x-20x=21-320$$

Add x to -20x.

$$-19x=21-320$$

Add 21 to -320.

$$-19x=-299$$

Divide both sides by -19.

$$x=\frac{299}{19}$$

Substitute \frac{299}{19} for x in 20x-y=320. Because the resulting equation contains only one variable, you can solve for y directly.

$$20\times \frac{299}{19}-y=320$$

Multiply 20 times \frac{299}{19}.

$$\frac{5980}{19}-y=320$$

Subtract \frac{5980}{19} from both sides of the equation.

$$-y=\frac{100}{19}$$

Divide both sides by -1.

$$y=-\frac{100}{19}$$

The system is now solved.

$$x=\frac{299}{19},y=-\frac{100}{19}$$