Solve {l}{5x-y/3=-6}{9x+2y/3=-7}

$$\left\{ \begin{array} { l } { 5 x - \frac { y } { 3 } = - 6 } \\ { 9 x + \frac { 2 y } { 3 } = - 7 } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=3$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 3.

$$15x-y=-18$$

Consider the second equation. Multiply both sides of the equation by 3.

$$27x+2y=-21$$

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.

$$15x-y=-18,27x+2y=-21$$

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

$$15x-y=-18$$

Add y to both sides of the equation.

$$15x=y-18$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(y-18\right)$$

Multiply \frac{1}{15} times y-18.

$$x=\frac{1}{15}y-\frac{6}{5}$$

Substitute \frac{y}{15}-\frac{6}{5} for x in the other equation, 27x+2y=-21.

$$27\left(\frac{1}{15}y-\frac{6}{5}\right)+2y=-21$$

Multiply 27 times \frac{y}{15}-\frac{6}{5}.

$$\frac{9}{5}y-\frac{162}{5}+2y=-21$$

Add \frac{9y}{5} to 2y.

$$\frac{19}{5}y-\frac{162}{5}=-21$$

Add \frac{162}{5} to both sides of the equation.

$$\frac{19}{5}y=\frac{57}{5}$$

Divide both sides of the equation by \frac{19}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=3$$

Substitute 3 for y in x=\frac{1}{15}y-\frac{6}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{15}\times 3-\frac{6}{5}$$

Multiply \frac{1}{15} times 3.

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

Add -\frac{6}{5} to \frac{1}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-1$$

The system is now solved.

$$x=-1,y=3$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 3.

$$15x-y=-18$$

Consider the second equation. Multiply both sides of the equation by 3.

$$27x+2y=-21$$

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

$$15x-y=-18,27x+2y=-21$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&-1\\27&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-18\\-21\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&-1\\27&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&-1\\27&2\end{matrix}\right))\left(\begin{matrix}15&-1\\27&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-1\\27&2\end{matrix}\right))\left(\begin{matrix}-18\\-21\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-1\\27&2\end{matrix}\right))\left(\begin{matrix}-18\\-21\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-1\\27&2\end{matrix}\right))\left(\begin{matrix}-18\\-21\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{15\times 2-\left(-27\right)}&-\frac{-1}{15\times 2-\left(-27\right)}\\-\frac{27}{15\times 2-\left(-27\right)}&\frac{15}{15\times 2-\left(-27\right)}\end{matrix}\right)\left(\begin{matrix}-18\\-21\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{57}&\frac{1}{57}\\-\frac{9}{19}&\frac{5}{19}\end{matrix}\right)\left(\begin{matrix}-18\\-21\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{57}\left(-18\right)+\frac{1}{57}\left(-21\right)\\-\frac{9}{19}\left(-18\right)+\frac{5}{19}\left(-21\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-1,y=3$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 3.

$$15x-y=-18$$

Consider the second equation. Multiply both sides of the equation by 3.

$$27x+2y=-21$$

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.

$$15x-y=-18,27x+2y=-21$$

To make 15x and 27x equal, multiply all terms on each side of the first equation by 27 and all terms on each side of the second by 15.

$$27\times 15x+27\left(-1\right)y=27\left(-18\right),15\times 27x+15\times 2y=15\left(-21\right)$$

Simplify.

$$405x-27y=-486,405x+30y=-315$$

Subtract 405x+30y=-315 from 405x-27y=-486 by subtracting like terms on each side of the equal sign.

$$405x-405x-27y-30y=-486+315$$

Add 405x to -405x. Terms 405x and -405x cancel out, leaving an equation with only one variable that can be solved.

$$-27y-30y=-486+315$$

Add -27y to -30y.

$$-57y=-486+315$$

Add -486 to 315.

$$-57y=-171$$

Divide both sides by -57.

$$y=3$$

Substitute 3 for y in 27x+2y=-21. Because the resulting equation contains only one variable, you can solve for x directly.

$$27x+2\times 3=-21$$

Multiply 2 times 3.

$$27x+6=-21$$

Subtract 6 from both sides of the equation.

$$27x=-27$$

Divide both sides by 27.

$$x=-1$$

The system is now solved.

$$x=-1,y=3$$