Solve {l}{x/5-y/6=-1/30}{x/3-frac{y}{20}=frac{13}{12}}

$$\left. \begin{array} { l } { \frac { x } { 5 } - \frac { y } { 6 } = - \frac { 1 } { 30 } } \\ { \frac { x } { 3 } - \frac { y } { 20 } = \frac { 13 } { 12 } } \end{array} \right.$$

Solve for x, y

$$x=4$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 5,6,30.

$$6x-5y=-1$$

Consider the second equation. Multiply both sides of the equation by 60, the least common multiple of 3,20,12.

$$20x-3y=65$$

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.

$$6x-5y=-1,20x-3y=65$$

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

$$6x-5y=-1$$

Add 5y to both sides of the equation.

$$6x=5y-1$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(5y-1\right)$$

Multiply \frac{1}{6} times 5y-1.

$$x=\frac{5}{6}y-\frac{1}{6}$$

Substitute \frac{5y-1}{6} for x in the other equation, 20x-3y=65.

$$20\left(\frac{5}{6}y-\frac{1}{6}\right)-3y=65$$

Multiply 20 times \frac{5y-1}{6}.

$$\frac{50}{3}y-\frac{10}{3}-3y=65$$

Add \frac{50y}{3} to -3y.

$$\frac{41}{3}y-\frac{10}{3}=65$$

Add \frac{10}{3} to both sides of the equation.

$$\frac{41}{3}y=\frac{205}{3}$$

Divide both sides of the equation by \frac{41}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=5$$

Substitute 5 for y in x=\frac{5}{6}y-\frac{1}{6}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{6}\times 5-\frac{1}{6}$$

Multiply \frac{5}{6} times 5.

$$x=\frac{25-1}{6}$$

Add -\frac{1}{6} to \frac{25}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=4$$

The system is now solved.

$$x=4,y=5$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 5,6,30.

$$6x-5y=-1$$

Consider the second equation. Multiply both sides of the equation by 60, the least common multiple of 3,20,12.

$$20x-3y=65$$

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

$$6x-5y=-1,20x-3y=65$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&-5\\20&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\65\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&-5\\20&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&-5\\20&-3\end{matrix}\right))\left(\begin{matrix}6&-5\\20&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-5\\20&-3\end{matrix}\right))\left(\begin{matrix}-1\\65\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-5\\20&-3\end{matrix}\right))\left(\begin{matrix}-1\\65\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-5\\20&-3\end{matrix}\right))\left(\begin{matrix}-1\\65\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{6\left(-3\right)-\left(-5\times 20\right)}&-\frac{-5}{6\left(-3\right)-\left(-5\times 20\right)}\\-\frac{20}{6\left(-3\right)-\left(-5\times 20\right)}&\frac{6}{6\left(-3\right)-\left(-5\times 20\right)}\end{matrix}\right)\left(\begin{matrix}-1\\65\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{82}&\frac{5}{82}\\-\frac{10}{41}&\frac{3}{41}\end{matrix}\right)\left(\begin{matrix}-1\\65\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{82}\left(-1\right)+\frac{5}{82}\times 65\\-\frac{10}{41}\left(-1\right)+\frac{3}{41}\times 65\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4,y=5$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 5,6,30.

$$6x-5y=-1$$

Consider the second equation. Multiply both sides of the equation by 60, the least common multiple of 3,20,12.

$$20x-3y=65$$

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.

$$6x-5y=-1,20x-3y=65$$

To make 6x and 20x equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 6.

$$20\times 6x+20\left(-5\right)y=20\left(-1\right),6\times 20x+6\left(-3\right)y=6\times 65$$

Simplify.

$$120x-100y=-20,120x-18y=390$$

Subtract 120x-18y=390 from 120x-100y=-20 by subtracting like terms on each side of the equal sign.

$$120x-120x-100y+18y=-20-390$$

Add 120x to -120x. Terms 120x and -120x cancel out, leaving an equation with only one variable that can be solved.

$$-100y+18y=-20-390$$

Add -100y to 18y.

$$-82y=-20-390$$

Add -20 to -390.

$$-82y=-410$$

Divide both sides by -82.

$$y=5$$

Substitute 5 for y in 20x-3y=65. Because the resulting equation contains only one variable, you can solve for x directly.

$$20x-3\times 5=65$$

Multiply -3 times 5.

$$20x-15=65$$

Add 15 to both sides of the equation.

$$20x=80$$

Divide both sides by 20.

$$x=4$$

The system is now solved.

$$x=4,y=5$$