Solve {l}{83v_{1}-33v_{2}=4000}{33v_{1}-63v_{2}=-800}

$$\left. \begin{array} { l } { 83 v _ { 1 } - 33 v _ { 2 } = 4000 } \\ { 33 v _ { 1 } - 63 v _ { 2 } = - 800 } \end{array} \right.$$

Solve for v_1, v_2

$$v_{1} = \frac{4640}{69} = 67\frac{17}{69} \approx 67.246376812$$

$$v_{2} = \frac{9920}{207} = 47\frac{191}{207} \approx 47.922705314$$

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.

$$83v_{1}-33v_{2}=4000,33v_{1}-63v_{2}=-800$$

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

$$83v_{1}-33v_{2}=4000$$

Add 33v_{2} to both sides of the equation.

$$83v_{1}=33v_{2}+4000$$

Divide both sides by 83.

$$v_{1}=\frac{1}{83}\left(33v_{2}+4000\right)$$

Multiply \frac{1}{83} times 33v_{2}+4000.

$$v_{1}=\frac{33}{83}v_{2}+\frac{4000}{83}$$

Substitute \frac{33v_{2}+4000}{83} for v_{1} in the other equation, 33v_{1}-63v_{2}=-800.

$$33\left(\frac{33}{83}v_{2}+\frac{4000}{83}\right)-63v_{2}=-800$$

Multiply 33 times \frac{33v_{2}+4000}{83}.

$$\frac{1089}{83}v_{2}+\frac{132000}{83}-63v_{2}=-800$$

Add \frac{1089v_{2}}{83} to -63v_{2}.

$$-\frac{4140}{83}v_{2}+\frac{132000}{83}=-800$$

Subtract \frac{132000}{83} from both sides of the equation.

$$-\frac{4140}{83}v_{2}=-\frac{198400}{83}$$

Divide both sides of the equation by -\frac{4140}{83}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$v_{2}=\frac{9920}{207}$$

Substitute \frac{9920}{207} for v_{2} in v_{1}=\frac{33}{83}v_{2}+\frac{4000}{83}. Because the resulting equation contains only one variable, you can solve for v_{1} directly.

$$v_{1}=\frac{33}{83}\times \frac{9920}{207}+\frac{4000}{83}$$

Multiply \frac{33}{83} times \frac{9920}{207} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$v_{1}=\frac{109120}{5727}+\frac{4000}{83}$$

Add \frac{4000}{83} to \frac{109120}{5727} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v_{1}=\frac{4640}{69}$$

The system is now solved.

$$v_{1}=\frac{4640}{69},v_{2}=\frac{9920}{207}$$

Steps Using Matrices

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

$$83v_{1}-33v_{2}=4000,33v_{1}-63v_{2}=-800$$

Write the equations in matrix form.

$$\left(\begin{matrix}83&-33\\33&-63\end{matrix}\right)\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=\left(\begin{matrix}4000\\-800\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}83&-33\\33&-63\end{matrix}\right).

$$inverse(\left(\begin{matrix}83&-33\\33&-63\end{matrix}\right))\left(\begin{matrix}83&-33\\33&-63\end{matrix}\right)\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}83&-33\\33&-63\end{matrix}\right))\left(\begin{matrix}4000\\-800\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}83&-33\\33&-63\end{matrix}\right))\left(\begin{matrix}4000\\-800\end{matrix}\right)$$

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

$$\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}83&-33\\33&-63\end{matrix}\right))\left(\begin{matrix}4000\\-800\end{matrix}\right)$$

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.

$$\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{63}{83\left(-63\right)-\left(-33\times 33\right)}&-\frac{-33}{83\left(-63\right)-\left(-33\times 33\right)}\\-\frac{33}{83\left(-63\right)-\left(-33\times 33\right)}&\frac{83}{83\left(-63\right)-\left(-33\times 33\right)}\end{matrix}\right)\left(\begin{matrix}4000\\-800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{7}{460}&-\frac{11}{1380}\\\frac{11}{1380}&-\frac{83}{4140}\end{matrix}\right)\left(\begin{matrix}4000\\-800\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{7}{460}\times 4000-\frac{11}{1380}\left(-800\right)\\\frac{11}{1380}\times 4000-\frac{83}{4140}\left(-800\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}v_{1}\\v_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{4640}{69}\\\frac{9920}{207}\end{matrix}\right)$$

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

$$v_{1}=\frac{4640}{69},v_{2}=\frac{9920}{207}$$

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.

$$83v_{1}-33v_{2}=4000,33v_{1}-63v_{2}=-800$$

To make 83v_{1} and 33v_{1} equal, multiply all terms on each side of the first equation by 33 and all terms on each side of the second by 83.

$$33\times 83v_{1}+33\left(-33\right)v_{2}=33\times 4000,83\times 33v_{1}+83\left(-63\right)v_{2}=83\left(-800\right)$$

Simplify.

$$2739v_{1}-1089v_{2}=132000,2739v_{1}-5229v_{2}=-66400$$

Subtract 2739v_{1}-5229v_{2}=-66400 from 2739v_{1}-1089v_{2}=132000 by subtracting like terms on each side of the equal sign.

$$2739v_{1}-2739v_{1}-1089v_{2}+5229v_{2}=132000+66400$$

Add 2739v_{1} to -2739v_{1}. Terms 2739v_{1} and -2739v_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-1089v_{2}+5229v_{2}=132000+66400$$

Add -1089v_{2} to 5229v_{2}.

$$4140v_{2}=132000+66400$$

Add 132000 to 66400.

$$4140v_{2}=198400$$

Divide both sides by 4140.

$$v_{2}=\frac{9920}{207}$$

Substitute \frac{9920}{207} for v_{2} in 33v_{1}-63v_{2}=-800. Because the resulting equation contains only one variable, you can solve for v_{1} directly.

$$33v_{1}-63\times \frac{9920}{207}=-800$$

Multiply -63 times \frac{9920}{207}.

$$33v_{1}-\frac{69440}{23}=-800$$

Add \frac{69440}{23} to both sides of the equation.

$$33v_{1}=\frac{51040}{23}$$

Divide both sides by 33.

$$v_{1}=\frac{4640}{69}$$

The system is now solved.

$$v_{1}=\frac{4640}{69},v_{2}=\frac{9920}{207}$$