Solve {l}{D1+D_{2}=60}{D_{1}=2/3D_{2}}

$$\left. \begin{array} { l } { D 1 + D _ { 2 } = 60 } \\ { D _ { 1 } = \frac { 2 } { 3 } D _ { 2 } } \end{array} \right.$$

Solve for D_1, D_2

$$D_{1}=24$$

$$D_{2}=36$$

Steps Using Substitution

Consider the second equation. Subtract \frac{2}{3}D_{2} from both sides.

$$D_{1}-\frac{2}{3}D_{2}=0$$

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.

$$D_{1}+D_{2}=60,D_{1}-\frac{2}{3}D_{2}=0$$

Choose one of the equations and solve it for D_{1} by isolating D_{1} on the left hand side of the equal sign.

$$D_{1}+D_{2}=60$$

Subtract D_{2} from both sides of the equation.

$$D_{1}=-D_{2}+60$$

Substitute -D_{2}+60 for D_{1} in the other equation, D_{1}-\frac{2}{3}D_{2}=0.

$$-D_{2}+60-\frac{2}{3}D_{2}=0$$

Add -D_{2} to -\frac{2D_{2}}{3}.

$$-\frac{5}{3}D_{2}+60=0$$

Subtract 60 from both sides of the equation.

$$-\frac{5}{3}D_{2}=-60$$

Divide both sides of the equation by -\frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$D_{2}=36$$

Substitute 36 for D_{2} in D_{1}=-D_{2}+60. Because the resulting equation contains only one variable, you can solve for D_{1} directly.

$$D_{1}=-36+60$$

Add 60 to -36.

$$D_{1}=24$$

The system is now solved.

$$D_{1}=24,D_{2}=36$$

Steps Using Matrices

Consider the second equation. Subtract \frac{2}{3}D_{2} from both sides.

$$D_{1}-\frac{2}{3}D_{2}=0$$

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

$$D_{1}+D_{2}=60,D_{1}-\frac{2}{3}D_{2}=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right)\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=\left(\begin{matrix}60\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right))\left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right)\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right))\left(\begin{matrix}60\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right))\left(\begin{matrix}60\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-\frac{2}{3}\end{matrix}\right))\left(\begin{matrix}60\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{2}{3}}{-\frac{2}{3}-1}&-\frac{1}{-\frac{2}{3}-1}\\-\frac{1}{-\frac{2}{3}-1}&\frac{1}{-\frac{2}{3}-1}\end{matrix}\right)\left(\begin{matrix}60\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}&\frac{3}{5}\\\frac{3}{5}&-\frac{3}{5}\end{matrix}\right)\left(\begin{matrix}60\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\times 60\\\frac{3}{5}\times 60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}D_{1}\\D_{2}\end{matrix}\right)=\left(\begin{matrix}24\\36\end{matrix}\right)$$

Extract the matrix elements D_{1} and D_{2}.

$$D_{1}=24,D_{2}=36$$

Steps Using Elimination

Consider the second equation. Subtract \frac{2}{3}D_{2} from both sides.

$$D_{1}-\frac{2}{3}D_{2}=0$$

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.

$$D_{1}+D_{2}=60,D_{1}-\frac{2}{3}D_{2}=0$$

Subtract D_{1}-\frac{2}{3}D_{2}=0 from D_{1}+D_{2}=60 by subtracting like terms on each side of the equal sign.

$$D_{1}-D_{1}+D_{2}+\frac{2}{3}D_{2}=60$$

Add D_{1} to -D_{1}. Terms D_{1} and -D_{1} cancel out, leaving an equation with only one variable that can be solved.

$$D_{2}+\frac{2}{3}D_{2}=60$$

Add D_{2} to \frac{2D_{2}}{3}.

$$\frac{5}{3}D_{2}=60$$

Divide both sides of the equation by \frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$D_{2}=36$$

Substitute 36 for D_{2} in D_{1}-\frac{2}{3}D_{2}=0. Because the resulting equation contains only one variable, you can solve for D_{1} directly.

$$D_{1}-\frac{2}{3}\times 36=0$$

Multiply -\frac{2}{3} times 36.

$$D_{1}-24=0$$

Add 24 to both sides of the equation.

$$D_{1}=24$$

The system is now solved.

$$D_{1}=24,D_{2}=36$$