Solve {l}{x+5*y-1/3=-2/3}{2*x+3/2-3*y=10,5}

$$\left\{ \begin{array} { l } { x + \frac { 5 \cdot y - 1 } { 3 } = - \frac { 2 } { 3 } } \\ { \frac { 2 \cdot x + 3 } { 2 } - 3 \cdot y = 10,5 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=-2$$

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.

$$x+\frac{1}{3}\left(5y-1\right)=-\frac{2}{3};\frac{1}{2}\left(2x+3\right)-3y=10,5$$

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

$$x+\frac{1}{3}\left(5y-1\right)=-\frac{2}{3}$$

Multiply \frac{1}{3} times 5y-1.

$$x+\frac{5}{3}y-\frac{1}{3}=-\frac{2}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x+\frac{5}{3}y=-\frac{1}{3}$$

Subtract \frac{5y}{3} from both sides of the equation.

$$x=-\frac{5}{3}y-\frac{1}{3}$$

Substitute \frac{-5y-1}{3} for x in the other equation, \frac{1}{2}\left(2x+3\right)-3y=10,5.

$$\frac{1}{2}\left(2\left(-\frac{5}{3}y-\frac{1}{3}\right)+3\right)-3y=10,5$$

Multiply 2 times \frac{-5y-1}{3}.

$$\frac{1}{2}\left(-\frac{10}{3}y-\frac{2}{3}+3\right)-3y=10,5$$

Add -\frac{2}{3} to 3.

$$\frac{1}{2}\left(-\frac{10}{3}y+\frac{7}{3}\right)-3y=10,5$$

Multiply \frac{1}{2} times \frac{-10y+7}{3}.

$$-\frac{5}{3}y+\frac{7}{6}-3y=10,5$$

Add -\frac{5y}{3} to -3y.

$$-\frac{14}{3}y+\frac{7}{6}=10,5$$

Subtract \frac{7}{6} from both sides of the equation.

$$-\frac{14}{3}y=\frac{28}{3}$$

Divide both sides of the equation by -\frac{14}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-2$$

Substitute -2 for y in x=-\frac{5}{3}y-\frac{1}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{3}\left(-2\right)-\frac{1}{3}$$

Multiply -\frac{5}{3} times -2.

$$x=\frac{10-1}{3}$$

Add -\frac{1}{3} to \frac{10}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=3$$

The system is now solved.

$$x=3;y=-2$$

Steps Using Matrices

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

$$x+\frac{1}{3}\left(5y-1\right)=-\frac{2}{3};\frac{1}{2}\left(2x+3\right)-3y=10,5$$

Simplify the first equation to put it in standard form.

$$x+\frac{1}{3}\left(5y-1\right)=-\frac{2}{3}$$

Multiply \frac{1}{3} times 5y-1.

$$x+\frac{5}{3}y-\frac{1}{3}=-\frac{2}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x+\frac{5}{3}y=-\frac{1}{3}$$

Simplify the second equation to put it in standard form.

$$\frac{1}{2}\left(2x+3\right)-3y=10,5$$

Multiply \frac{1}{2} times 2x+3.

$$x+\frac{3}{2}-3y=10,5$$

Subtract \frac{3}{2} from both sides of the equation.

$$x-3y=9$$

Write the equations in matrix form.

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

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&\frac{5}{3}\\1&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&\frac{5}{3}\\1&-3\end{matrix}\right))\left(\begin{matrix}1&\frac{5}{3}\\1&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&\frac{5}{3}\\1&-3\end{matrix}\right))\left(\begin{matrix}-\frac{1}{3}\\9\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&\frac{5}{3}\\1&-3\end{matrix}\right))\left(\begin{matrix}-\frac{1}{3}\\9\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&\frac{5}{3}\\1&-3\end{matrix}\right))\left(\begin{matrix}-\frac{1}{3}\\9\end{matrix}\right)$$

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.

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

Do the arithmetic.

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

Multiply the matrices.

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

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\-2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3;y=-2$$