Solve {c}{x-2y=3}{5x-3(y+2)=2}

$$\left\{ \begin{array} { c } { x - 2 y = 3 } \\ { 5 x - 3 ( y + 2 ) = 2 } \end{array} \right.$$

Solve for x, y

$$x=1$$

$$y=-1$$

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.

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

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

$$x-2y=3$$

Add 2y to both sides of the equation.

$$x=2y+3$$

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

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

Multiply 5 times 2y+3.

$$10y+15-3\left(y+2\right)=2$$

Multiply -3 times y+2.

$$10y+15-3y-6=2$$

Add 10y to -3y.

$$7y+15-6=2$$

Add 15 to -6.

$$7y+9=2$$

Subtract 9 from both sides of the equation.

$$7y=-7$$

Divide both sides by 7.

$$y=-1$$

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

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

Multiply 2 times -1.

$$x=-2+3$$

Add 3 to -2.

$$x=1$$

The system is now solved.

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

Steps Using Matrices

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

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

Simplify the second equation to put it in standard form.

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

Multiply -3 times y+2.

$$5x-3y-6=2$$

Add 6 to both sides of the equation.

$$5x-3y=8$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2\\5&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\8\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&-2\\5&-3\end{matrix}\right))\left(\begin{matrix}1&-2\\5&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\5&-3\end{matrix}\right))\left(\begin{matrix}3\\8\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\5&-3\end{matrix}\right))\left(\begin{matrix}3\\8\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\5&-3\end{matrix}\right))\left(\begin{matrix}3\\8\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-\left(-2\times 5\right)}&-\frac{-2}{-3-\left(-2\times 5\right)}\\-\frac{5}{-3-\left(-2\times 5\right)}&\frac{1}{-3-\left(-2\times 5\right)}\end{matrix}\right)\left(\begin{matrix}3\\8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{7}&\frac{2}{7}\\-\frac{5}{7}&\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}3\\8\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{7}\times 3+\frac{2}{7}\times 8\\-\frac{5}{7}\times 3+\frac{1}{7}\times 8\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

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