Solve {l}{2(x+2)-3(y-1)=13}{3(x+2)+5(y-1)=30.9}

$$\left\{ \begin{array} { l } { 2 ( x + 2 ) - 3 ( y - 1 ) = 13 } \\ { 3 ( x + 2 ) + 5 ( y - 1 ) = 30.9 } \end{array} \right.$$

Solve for x, y

$$x=6.3$$

$$y=2.2$$

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.

$$2\left(x+2\right)-3\left(y-1\right)=13,3\left(x+2\right)+5\left(y-1\right)=30.9$$

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

$$2\left(x+2\right)-3\left(y-1\right)=13$$

Multiply 2 times x+2.

$$2x+4-3\left(y-1\right)=13$$

Multiply -3 times y-1.

$$2x+4-3y+3=13$$

Add 4 to 3.

$$2x-3y+7=13$$

Subtract 7 from both sides of the equation.

$$2x-3y=6$$

Add 3y to both sides of the equation.

$$2x=3y+6$$

Divide both sides by 2.

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

Multiply \frac{1}{2} times 6+3y.

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

Substitute \frac{3y}{2}+3 for x in the other equation, 3\left(x+2\right)+5\left(y-1\right)=30.9.

$$3\left(\frac{3}{2}y+3+2\right)+5\left(y-1\right)=30.9$$

Add 3 to 2.

$$3\left(\frac{3}{2}y+5\right)+5\left(y-1\right)=30.9$$

Multiply 3 times \frac{3y}{2}+5.

$$\frac{9}{2}y+15+5\left(y-1\right)=30.9$$

Multiply 5 times y-1.

$$\frac{9}{2}y+15+5y-5=30.9$$

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

$$\frac{19}{2}y+15-5=30.9$$

Add 15 to -5.

$$\frac{19}{2}y+10=30.9$$

Subtract 10 from both sides of the equation.

$$\frac{19}{2}y=20.9$$

Divide both sides of the equation by \frac{19}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{11}{5}$$

Substitute \frac{11}{5} for y in x=\frac{3}{2}y+3. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{2}\times \frac{11}{5}+3$$

Multiply \frac{3}{2} times \frac{11}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{33}{10}+3$$

Add 3 to \frac{33}{10}.

$$x=\frac{63}{10}$$

The system is now solved.

$$x=\frac{63}{10},y=\frac{11}{5}$$

Steps Using Matrices

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

$$2\left(x+2\right)-3\left(y-1\right)=13,3\left(x+2\right)+5\left(y-1\right)=30.9$$

Simplify the first equation to put it in standard form.

$$2\left(x+2\right)-3\left(y-1\right)=13$$

Multiply 2 times x+2.

$$2x+4-3\left(y-1\right)=13$$

Multiply -3 times y-1.

$$2x+4-3y+3=13$$

Add 4 to 3.

$$2x-3y+7=13$$

Subtract 7 from both sides of the equation.

$$2x-3y=6$$

Simplify the second equation to put it in standard form.

$$3\left(x+2\right)+5\left(y-1\right)=30.9$$

Multiply 3 times x+2.

$$3x+6+5\left(y-1\right)=30.9$$

Multiply 5 times y-1.

$$3x+6+5y-5=30.9$$

Add 6 to -5.

$$3x+5y+1=30.9$$

Subtract 1 from both sides of the equation.

$$3x+5y=29.9$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\29.9\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}2&-3\\3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}6\\29.9\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}6\\29.9\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\3&5\end{matrix}\right))\left(\begin{matrix}6\\29.9\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{2\times 5-\left(-3\times 3\right)}&-\frac{-3}{2\times 5-\left(-3\times 3\right)}\\-\frac{3}{2\times 5-\left(-3\times 3\right)}&\frac{2}{2\times 5-\left(-3\times 3\right)}\end{matrix}\right)\left(\begin{matrix}6\\29.9\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}&\frac{3}{19}\\-\frac{3}{19}&\frac{2}{19}\end{matrix}\right)\left(\begin{matrix}6\\29.9\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}\times 6+\frac{3}{19}\times 29.9\\-\frac{3}{19}\times 6+\frac{2}{19}\times 29.9\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{63}{10}\\\frac{11}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{63}{10},y=\frac{11}{5}$$