Solve {l}{5(x-4)+3(y-2)=1}{x+1/3+y+4/2=23/6}

$$\left\{ \begin{array} { l } { 5 ( x - 4 ) + 3 ( y - 2 ) = 1 } \\ { \frac { x + 1 } { 3 } + \frac { y + 4 } { 2 } = \frac { 23 } { 6 } } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=-1$$

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.

$$5\left(x-4\right)+3\left(y-2\right)=1,\frac{1}{3}\left(x+1\right)+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

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

$$5\left(x-4\right)+3\left(y-2\right)=1$$

Multiply 5 times x-4.

$$5x-20+3\left(y-2\right)=1$$

Multiply 3 times y-2.

$$5x-20+3y-6=1$$

Add -20 to -6.

$$5x+3y-26=1$$

Add 26 to both sides of the equation.

$$5x+3y=27$$

Subtract 3y from both sides of the equation.

$$5x=-3y+27$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-3y+27\right)$$

Multiply \frac{1}{5} times -3y+27.

$$x=-\frac{3}{5}y+\frac{27}{5}$$

Substitute \frac{-3y+27}{5} for x in the other equation, \frac{1}{3}\left(x+1\right)+\frac{1}{2}\left(y+4\right)=\frac{23}{6}.

$$\frac{1}{3}\left(-\frac{3}{5}y+\frac{27}{5}+1\right)+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

Add \frac{27}{5} to 1.

$$\frac{1}{3}\left(-\frac{3}{5}y+\frac{32}{5}\right)+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

Multiply \frac{1}{3} times \frac{-3y+32}{5}.

$$-\frac{1}{5}y+\frac{32}{15}+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

Multiply \frac{1}{2} times y+4.

$$-\frac{1}{5}y+\frac{32}{15}+\frac{1}{2}y+2=\frac{23}{6}$$

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

$$\frac{3}{10}y+\frac{32}{15}+2=\frac{23}{6}$$

Add \frac{32}{15} to 2.

$$\frac{3}{10}y+\frac{62}{15}=\frac{23}{6}$$

Subtract \frac{62}{15} from both sides of the equation.

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

Divide both sides of the equation by \frac{3}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-1$$

Substitute -1 for y in x=-\frac{3}{5}y+\frac{27}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{5}\left(-1\right)+\frac{27}{5}$$

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

$$x=\frac{3+27}{5}$$

Add \frac{27}{5} to \frac{3}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=6$$

The system is now solved.

$$x=6,y=-1$$

Steps Using Matrices

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

$$5\left(x-4\right)+3\left(y-2\right)=1,\frac{1}{3}\left(x+1\right)+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

Simplify the first equation to put it in standard form.

$$5\left(x-4\right)+3\left(y-2\right)=1$$

Multiply 5 times x-4.

$$5x-20+3\left(y-2\right)=1$$

Multiply 3 times y-2.

$$5x-20+3y-6=1$$

Add -20 to -6.

$$5x+3y-26=1$$

Add 26 to both sides of the equation.

$$5x+3y=27$$

Simplify the second equation to put it in standard form.

$$\frac{1}{3}\left(x+1\right)+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

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

$$\frac{1}{3}x+\frac{1}{3}+\frac{1}{2}\left(y+4\right)=\frac{23}{6}$$

Multiply \frac{1}{2} times y+4.

$$\frac{1}{3}x+\frac{1}{3}+\frac{1}{2}y+2=\frac{23}{6}$$

Add \frac{1}{3} to 2.

$$\frac{1}{3}x+\frac{1}{2}y+\frac{7}{3}=\frac{23}{6}$$

Subtract \frac{7}{3} from both sides of the equation.

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

Write the equations in matrix form.

$$\left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}27\\\frac{3}{2}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}27\\\frac{3}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}27\\\frac{3}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&3\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}27\\\frac{3}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{2}}{5\times \frac{1}{2}-3\times \frac{1}{3}}&-\frac{3}{5\times \frac{1}{2}-3\times \frac{1}{3}}\\-\frac{\frac{1}{3}}{5\times \frac{1}{2}-3\times \frac{1}{3}}&\frac{5}{5\times \frac{1}{2}-3\times \frac{1}{3}}\end{matrix}\right)\left(\begin{matrix}27\\\frac{3}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}&-2\\-\frac{2}{9}&\frac{10}{3}\end{matrix}\right)\left(\begin{matrix}27\\\frac{3}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 27-2\times \frac{3}{2}\\-\frac{2}{9}\times 27+\frac{10}{3}\times \frac{3}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\-1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=-1$$