Solve {l}{2x+5y+289=15}{3y+19x+336=18}

$$\left\{ \begin{array} { l } { 2 x + 5 y + 289 = 15 } \\ { 3 y + 19 x + 336 = 18 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{768}{89} = -8\frac{56}{89} \approx -8.629213483$$

$$y = -\frac{4570}{89} = -51\frac{31}{89} \approx -51.348314607$$

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.

$$2x+5y+289=15,19x+3y+336=18$$

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

$$2x+5y+289=15$$

Subtract 289 from both sides of the equation.

$$2x+5y=-274$$

Subtract 5y from both sides of the equation.

$$2x=-5y-274$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-5y-274\right)$$

Multiply \frac{1}{2} times -5y-274.

$$x=-\frac{5}{2}y-137$$

Substitute -\frac{5y}{2}-137 for x in the other equation, 19x+3y+336=18.

$$19\left(-\frac{5}{2}y-137\right)+3y+336=18$$

Multiply 19 times -\frac{5y}{2}-137.

$$-\frac{95}{2}y-2603+3y+336=18$$

Add -\frac{95y}{2} to 3y.

$$-\frac{89}{2}y-2603+336=18$$

Add -2603 to 336.

$$-\frac{89}{2}y-2267=18$$

Add 2267 to both sides of the equation.

$$-\frac{89}{2}y=2285$$

Divide both sides of the equation by -\frac{89}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{4570}{89}$$

Substitute -\frac{4570}{89} for y in x=-\frac{5}{2}y-137. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{2}\left(-\frac{4570}{89}\right)-137$$

Multiply -\frac{5}{2} times -\frac{4570}{89} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{11425}{89}-137$$

Add -137 to \frac{11425}{89}.

$$x=-\frac{768}{89}$$

The system is now solved.

$$x=-\frac{768}{89},y=-\frac{4570}{89}$$

Steps Using Matrices

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

$$2x+5y+289=15,19x+3y+336=18$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&5\\19&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-274\\-318\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&5\\19&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&5\\19&3\end{matrix}\right))\left(\begin{matrix}2&5\\19&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&5\\19&3\end{matrix}\right))\left(\begin{matrix}-274\\-318\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&5\\19&3\end{matrix}\right))\left(\begin{matrix}-274\\-318\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&5\\19&3\end{matrix}\right))\left(\begin{matrix}-274\\-318\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{2\times 3-5\times 19}&-\frac{5}{2\times 3-5\times 19}\\-\frac{19}{2\times 3-5\times 19}&\frac{2}{2\times 3-5\times 19}\end{matrix}\right)\left(\begin{matrix}-274\\-318\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{89}&\frac{5}{89}\\\frac{19}{89}&-\frac{2}{89}\end{matrix}\right)\left(\begin{matrix}-274\\-318\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{89}\left(-274\right)+\frac{5}{89}\left(-318\right)\\\frac{19}{89}\left(-274\right)-\frac{2}{89}\left(-318\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{768}{89}\\-\frac{4570}{89}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{768}{89},y=-\frac{4570}{89}$$

Steps Using Elimination

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.

$$2x+5y+289=15,19x+3y+336=18$$

To make 2x and 19x equal, multiply all terms on each side of the first equation by 19 and all terms on each side of the second by 2.

$$19\times 2x+19\times 5y+19\times 289=19\times 15,2\times 19x+2\times 3y+2\times 336=2\times 18$$

Simplify.

$$38x+95y+5491=285,38x+6y+672=36$$

Subtract 38x+6y+672=36 from 38x+95y+5491=285 by subtracting like terms on each side of the equal sign.

$$38x-38x+95y-6y+5491-672=285-36$$

Add 38x to -38x. Terms 38x and -38x cancel out, leaving an equation with only one variable that can be solved.

$$95y-6y+5491-672=285-36$$

Add 95y to -6y.

$$89y+5491-672=285-36$$

Add 5491 to -672.

$$89y+4819=285-36$$

Add 285 to -36.

$$89y+4819=249$$

Subtract 4819 from both sides of the equation.

$$89y=-4570$$

Divide both sides by 89.

$$y=-\frac{4570}{89}$$

Substitute -\frac{4570}{89} for y in 19x+3y+336=18. Because the resulting equation contains only one variable, you can solve for x directly.

$$19x+3\left(-\frac{4570}{89}\right)+336=18$$

Multiply 3 times -\frac{4570}{89}.

$$19x-\frac{13710}{89}+336=18$$

Add -\frac{13710}{89} to 336.

$$19x+\frac{16194}{89}=18$$

Subtract \frac{16194}{89} from both sides of the equation.

$$19x=-\frac{14592}{89}$$

Divide both sides by 19.

$$x=-\frac{768}{89}$$

The system is now solved.

$$x=-\frac{768}{89},y=-\frac{4570}{89}$$