Solve {l}{2x+y=36}{400x+360y=500x+450y-1960}

$$\left\{ \begin{array} { l } { 2 x + y = 36 } \\ { 400 x + 360 y = 500 x + 450 y - 1960 } \end{array} \right.$$

Solve for x, y

$$x=16$$

$$y=4$$

Steps Using Substitution

Consider the second equation. Subtract 500x from both sides.

$$400x+360y-500x=450y-1960$$

Combine 400x and -500x to get -100x.

$$-100x+360y=450y-1960$$

Subtract 450y from both sides.

$$-100x+360y-450y=-1960$$

Combine 360y and -450y to get -90y.

$$-100x-90y=-1960$$

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.

$$2x+y=36,-100x-90y=-1960$$

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

$$2x+y=36$$

Subtract y from both sides of the equation.

$$2x=-y+36$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-y+36\right)$$

Multiply \frac{1}{2} times -y+36.

$$x=-\frac{1}{2}y+18$$

Substitute -\frac{y}{2}+18 for x in the other equation, -100x-90y=-1960.

$$-100\left(-\frac{1}{2}y+18\right)-90y=-1960$$

Multiply -100 times -\frac{y}{2}+18.

$$50y-1800-90y=-1960$$

Add 50y to -90y.

$$-40y-1800=-1960$$

Add 1800 to both sides of the equation.

$$-40y=-160$$

Divide both sides by -40.

$$y=4$$

Substitute 4 for y in x=-\frac{1}{2}y+18. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{2}\times 4+18$$

Multiply -\frac{1}{2} times 4.

$$x=-2+18$$

Add 18 to -2.

$$x=16$$

The system is now solved.

$$x=16,y=4$$

Steps Using Matrices

Consider the second equation. Subtract 500x from both sides.

$$400x+360y-500x=450y-1960$$

Combine 400x and -500x to get -100x.

$$-100x+360y=450y-1960$$

Subtract 450y from both sides.

$$-100x+360y-450y=-1960$$

Combine 360y and -450y to get -90y.

$$-100x-90y=-1960$$

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

$$2x+y=36,-100x-90y=-1960$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&1\\-100&-90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36\\-1960\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&1\\-100&-90\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&1\\-100&-90\end{matrix}\right))\left(\begin{matrix}2&1\\-100&-90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\-100&-90\end{matrix}\right))\left(\begin{matrix}36\\-1960\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\-100&-90\end{matrix}\right))\left(\begin{matrix}36\\-1960\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\-100&-90\end{matrix}\right))\left(\begin{matrix}36\\-1960\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{90}{2\left(-90\right)-\left(-100\right)}&-\frac{1}{2\left(-90\right)-\left(-100\right)}\\-\frac{-100}{2\left(-90\right)-\left(-100\right)}&\frac{2}{2\left(-90\right)-\left(-100\right)}\end{matrix}\right)\left(\begin{matrix}36\\-1960\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{9}{8}&\frac{1}{80}\\-\frac{5}{4}&-\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}36\\-1960\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{9}{8}\times 36+\frac{1}{80}\left(-1960\right)\\-\frac{5}{4}\times 36-\frac{1}{40}\left(-1960\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}16\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=16,y=4$$

Steps Using Elimination

Consider the second equation. Subtract 500x from both sides.

$$400x+360y-500x=450y-1960$$

Combine 400x and -500x to get -100x.

$$-100x+360y=450y-1960$$

Subtract 450y from both sides.

$$-100x+360y-450y=-1960$$

Combine 360y and -450y to get -90y.

$$-100x-90y=-1960$$

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.

$$2x+y=36,-100x-90y=-1960$$

To make 2x and -100x equal, multiply all terms on each side of the first equation by -100 and all terms on each side of the second by 2.

$$-100\times 2x-100y=-100\times 36,2\left(-100\right)x+2\left(-90\right)y=2\left(-1960\right)$$

Simplify.

$$-200x-100y=-3600,-200x-180y=-3920$$

Subtract -200x-180y=-3920 from -200x-100y=-3600 by subtracting like terms on each side of the equal sign.

$$-200x+200x-100y+180y=-3600+3920$$

Add -200x to 200x. Terms -200x and 200x cancel out, leaving an equation with only one variable that can be solved.

$$-100y+180y=-3600+3920$$

Add -100y to 180y.

$$80y=-3600+3920$$

Add -3600 to 3920.

$$80y=320$$

Divide both sides by 80.

$$y=4$$

Substitute 4 for y in -100x-90y=-1960. Because the resulting equation contains only one variable, you can solve for x directly.

$$-100x-90\times 4=-1960$$

Multiply -90 times 4.

$$-100x-360=-1960$$

Add 360 to both sides of the equation.

$$-100x=-1600$$

Divide both sides by -100.

$$x=16$$

The system is now solved.

$$x=16,y=4$$