Solve {l}{12x_{1}+7x_{2}=147}{15x_{1}+19x_{2}=168}

$$\left. \begin{array}{l}{ 12 x _ { 1 } + 7 x _ { 2 } = 147 }\\{ 15 x _ { 1 } + 19 x _ { 2 } = 168 }\end{array} \right.$$

Solve for x_1, x_2

$$x_{1} = \frac{539}{41} = 13\frac{6}{41} \approx 13.146341463$$

$$x_{2} = -\frac{63}{41} = -1\frac{22}{41} \approx -1.536585366$$

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.

$$12x_{1}+7x_{2}=147,15x_{1}+19x_{2}=168$$

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

$$12x_{1}+7x_{2}=147$$

Subtract 7x_{2} from both sides of the equation.

$$12x_{1}=-7x_{2}+147$$

Divide both sides by 12.

$$x_{1}=\frac{1}{12}\left(-7x_{2}+147\right)$$

Multiply \frac{1}{12} times -7x_{2}+147.

$$x_{1}=-\frac{7}{12}x_{2}+\frac{49}{4}$$

Substitute -\frac{7x_{2}}{12}+\frac{49}{4} for x_{1} in the other equation, 15x_{1}+19x_{2}=168.

$$15\left(-\frac{7}{12}x_{2}+\frac{49}{4}\right)+19x_{2}=168$$

Multiply 15 times -\frac{7x_{2}}{12}+\frac{49}{4}.

$$-\frac{35}{4}x_{2}+\frac{735}{4}+19x_{2}=168$$

Add -\frac{35x_{2}}{4} to 19x_{2}.

$$\frac{41}{4}x_{2}+\frac{735}{4}=168$$

Subtract \frac{735}{4} from both sides of the equation.

$$\frac{41}{4}x_{2}=-\frac{63}{4}$$

Divide both sides of the equation by \frac{41}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x_{2}=-\frac{63}{41}$$

Substitute -\frac{63}{41} for x_{2} in x_{1}=-\frac{7}{12}x_{2}+\frac{49}{4}. Because the resulting equation contains only one variable, you can solve for x_{1} directly.

$$x_{1}=-\frac{7}{12}\left(-\frac{63}{41}\right)+\frac{49}{4}$$

Multiply -\frac{7}{12} times -\frac{63}{41} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x_{1}=\frac{147}{164}+\frac{49}{4}$$

Add \frac{49}{4} to \frac{147}{164} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x_{1}=\frac{539}{41}$$

The system is now solved.

$$x_{1}=\frac{539}{41},x_{2}=-\frac{63}{41}$$

Steps Using Matrices

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

$$12x_{1}+7x_{2}=147,15x_{1}+19x_{2}=168$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&7\\15&19\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}147\\168\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&7\\15&19\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&7\\15&19\end{matrix}\right))\left(\begin{matrix}12&7\\15&19\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}12&7\\15&19\end{matrix}\right))\left(\begin{matrix}147\\168\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}12&7\\15&19\end{matrix}\right))\left(\begin{matrix}147\\168\end{matrix}\right)$$

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

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}12&7\\15&19\end{matrix}\right))\left(\begin{matrix}147\\168\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{19}{12\times 19-7\times 15}&-\frac{7}{12\times 19-7\times 15}\\-\frac{15}{12\times 19-7\times 15}&\frac{12}{12\times 19-7\times 15}\end{matrix}\right)\left(\begin{matrix}147\\168\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{19}{123}&-\frac{7}{123}\\-\frac{5}{41}&\frac{4}{41}\end{matrix}\right)\left(\begin{matrix}147\\168\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{19}{123}\times 147-\frac{7}{123}\times 168\\-\frac{5}{41}\times 147+\frac{4}{41}\times 168\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{539}{41}\\-\frac{63}{41}\end{matrix}\right)$$

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

$$x_{1}=\frac{539}{41},x_{2}=-\frac{63}{41}$$

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.

$$12x_{1}+7x_{2}=147,15x_{1}+19x_{2}=168$$

To make 12x_{1} and 15x_{1} equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 12.

$$15\times 12x_{1}+15\times 7x_{2}=15\times 147,12\times 15x_{1}+12\times 19x_{2}=12\times 168$$

Simplify.

$$180x_{1}+105x_{2}=2205,180x_{1}+228x_{2}=2016$$

Subtract 180x_{1}+228x_{2}=2016 from 180x_{1}+105x_{2}=2205 by subtracting like terms on each side of the equal sign.

$$180x_{1}-180x_{1}+105x_{2}-228x_{2}=2205-2016$$

Add 180x_{1} to -180x_{1}. Terms 180x_{1} and -180x_{1} cancel out, leaving an equation with only one variable that can be solved.

$$105x_{2}-228x_{2}=2205-2016$$

Add 105x_{2} to -228x_{2}.

$$-123x_{2}=2205-2016$$

Add 2205 to -2016.

$$-123x_{2}=189$$

Divide both sides by -123.

$$x_{2}=-\frac{63}{41}$$

Substitute -\frac{63}{41} for x_{2} in 15x_{1}+19x_{2}=168. Because the resulting equation contains only one variable, you can solve for x_{1} directly.

$$15x_{1}+19\left(-\frac{63}{41}\right)=168$$

Multiply 19 times -\frac{63}{41}.

$$15x_{1}-\frac{1197}{41}=168$$

Add \frac{1197}{41} to both sides of the equation.

$$15x_{1}=\frac{8085}{41}$$

Divide both sides by 15.

$$x_{1}=\frac{539}{41}$$

The system is now solved.

$$x_{1}=\frac{539}{41},x_{2}=-\frac{63}{41}$$