Solve {l}{3x/7-2y/13=5}{x+1/3y=3/5}

$$\left. \begin{array} { l } { \frac { 3 x } { 7 } - \frac { 2 y } { 13 } = 5 } \\ { x + \frac { 1 } { 3 } y = \frac { 3 } { 5 } } \end{array} \right.$$

Solve for x, y

$$x = \frac{2401}{405} = 5\frac{376}{405} \approx 5.928395062$$

$$y = -\frac{2158}{135} = -15\frac{133}{135} \approx -15.985185185$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 91, the least common multiple of 7,13.

$$13\times 3x-7\times 2y=455$$

Multiply 13 and 3 to get 39.

$$39x-7\times 2y=455$$

Multiply -7 and 2 to get -14.

$$39x-14y=455$$

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.

$$39x-14y=455,x+\frac{1}{3}y=\frac{3}{5}$$

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

$$39x-14y=455$$

Add 14y to both sides of the equation.

$$39x=14y+455$$

Divide both sides by 39.

$$x=\frac{1}{39}\left(14y+455\right)$$

Multiply \frac{1}{39} times 14y+455.

$$x=\frac{14}{39}y+\frac{35}{3}$$

Substitute \frac{14y}{39}+\frac{35}{3} for x in the other equation, x+\frac{1}{3}y=\frac{3}{5}.

$$\frac{14}{39}y+\frac{35}{3}+\frac{1}{3}y=\frac{3}{5}$$

Add \frac{14y}{39} to \frac{y}{3}.

$$\frac{9}{13}y+\frac{35}{3}=\frac{3}{5}$$

Subtract \frac{35}{3} from both sides of the equation.

$$\frac{9}{13}y=-\frac{166}{15}$$

Divide both sides of the equation by \frac{9}{13}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{2158}{135}$$

Substitute -\frac{2158}{135} for y in x=\frac{14}{39}y+\frac{35}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{14}{39}\left(-\frac{2158}{135}\right)+\frac{35}{3}$$

Multiply \frac{14}{39} times -\frac{2158}{135} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{2324}{405}+\frac{35}{3}$$

Add \frac{35}{3} to -\frac{2324}{405} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{2401}{405}$$

The system is now solved.

$$x=\frac{2401}{405},y=-\frac{2158}{135}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 91, the least common multiple of 7,13.

$$13\times 3x-7\times 2y=455$$

Multiply 13 and 3 to get 39.

$$39x-7\times 2y=455$$

Multiply -7 and 2 to get -14.

$$39x-14y=455$$

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

$$39x-14y=455,x+\frac{1}{3}y=\frac{3}{5}$$

Write the equations in matrix form.

$$\left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}455\\\frac{3}{5}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right).

$$inverse(\left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}455\\\frac{3}{5}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}455\\\frac{3}{5}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}39&-14\\1&\frac{1}{3}\end{matrix}\right))\left(\begin{matrix}455\\\frac{3}{5}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{3}}{39\times \frac{1}{3}-\left(-14\right)}&-\frac{-14}{39\times \frac{1}{3}-\left(-14\right)}\\-\frac{1}{39\times \frac{1}{3}-\left(-14\right)}&\frac{39}{39\times \frac{1}{3}-\left(-14\right)}\end{matrix}\right)\left(\begin{matrix}455\\\frac{3}{5}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{81}&\frac{14}{27}\\-\frac{1}{27}&\frac{13}{9}\end{matrix}\right)\left(\begin{matrix}455\\\frac{3}{5}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{81}\times 455+\frac{14}{27}\times \frac{3}{5}\\-\frac{1}{27}\times 455+\frac{13}{9}\times \frac{3}{5}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2401}{405}\\-\frac{2158}{135}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{2401}{405},y=-\frac{2158}{135}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 91, the least common multiple of 7,13.

$$13\times 3x-7\times 2y=455$$

Multiply 13 and 3 to get 39.

$$39x-7\times 2y=455$$

Multiply -7 and 2 to get -14.

$$39x-14y=455$$

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.

$$39x-14y=455,x+\frac{1}{3}y=\frac{3}{5}$$

To make 39x 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 39.

$$39x-14y=455,39x+39\times \frac{1}{3}y=39\times \frac{3}{5}$$

Simplify.

$$39x-14y=455,39x+13y=\frac{117}{5}$$

Subtract 39x+13y=\frac{117}{5} from 39x-14y=455 by subtracting like terms on each side of the equal sign.

$$39x-39x-14y-13y=455-\frac{117}{5}$$

Add 39x to -39x. Terms 39x and -39x cancel out, leaving an equation with only one variable that can be solved.

$$-14y-13y=455-\frac{117}{5}$$

Add -14y to -13y.

$$-27y=455-\frac{117}{5}$$

Add 455 to -\frac{117}{5}.

$$-27y=\frac{2158}{5}$$

Divide both sides by -27.

$$y=-\frac{2158}{135}$$

Substitute -\frac{2158}{135} for y in x+\frac{1}{3}y=\frac{3}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+\frac{1}{3}\left(-\frac{2158}{135}\right)=\frac{3}{5}$$

Multiply \frac{1}{3} times -\frac{2158}{135} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x-\frac{2158}{405}=\frac{3}{5}$$

Add \frac{2158}{405} to both sides of the equation.

$$x=\frac{2401}{405}$$

The system is now solved.

$$x=\frac{2401}{405},y=-\frac{2158}{135}$$