Solve {l}{2*25x=40y}{x+y=36}

$$\left\{ \begin{array} { l } { 2 \times 25 x = 40 y } \\ { x + y = 36 } \end{array} \right.$$

Solve for x, y

$$x=16$$

$$y=20$$

Steps Using Substitution

Consider the first equation. Subtract 40y from both sides.

$$2\times 25x-40y=0$$

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.

$$50x-40y=0,x+y=36$$

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

$$50x-40y=0$$

Add 40y to both sides of the equation.

$$50x=40y$$

Divide both sides by 50.

$$x=\frac{1}{50}\times 40y$$

Multiply \frac{1}{50} times 40y.

$$x=\frac{4}{5}y$$

Substitute \frac{4y}{5} for x in the other equation, x+y=36.

$$\frac{4}{5}y+y=36$$

Add \frac{4y}{5} to y.

$$\frac{9}{5}y=36$$

Divide both sides of the equation by \frac{9}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=20$$

Substitute 20 for y in x=\frac{4}{5}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{5}\times 20$$

Multiply \frac{4}{5} times 20.

$$x=16$$

The system is now solved.

$$x=16,y=20$$

Steps Using Matrices

Consider the first equation. Subtract 40y from both sides.

$$2\times 25x-40y=0$$

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

$$50x-40y=0,x+y=36$$

Write the equations in matrix form.

$$\left(\begin{matrix}50&-40\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\36\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}50&-40\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}50&-40\\1&1\end{matrix}\right))\left(\begin{matrix}50&-40\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&-40\\1&1\end{matrix}\right))\left(\begin{matrix}0\\36\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&-40\\1&1\end{matrix}\right))\left(\begin{matrix}0\\36\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&-40\\1&1\end{matrix}\right))\left(\begin{matrix}0\\36\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{50-\left(-40\right)}&-\frac{-40}{50-\left(-40\right)}\\-\frac{1}{50-\left(-40\right)}&\frac{50}{50-\left(-40\right)}\end{matrix}\right)\left(\begin{matrix}0\\36\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{90}&\frac{4}{9}\\-\frac{1}{90}&\frac{5}{9}\end{matrix}\right)\left(\begin{matrix}0\\36\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{9}\times 36\\\frac{5}{9}\times 36\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}16\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=16,y=20$$

Steps Using Elimination

Consider the first equation. Subtract 40y from both sides.

$$2\times 25x-40y=0$$

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.

$$50x-40y=0,x+y=36$$

To make 50x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 50.

$$50x-40y=0,50x+50y=50\times 36$$

Simplify.

$$50x-40y=0,50x+50y=1800$$

Subtract 50x+50y=1800 from 50x-40y=0 by subtracting like terms on each side of the equal sign.

$$50x-50x-40y-50y=-1800$$

Add 50x to -50x. Terms 50x and -50x cancel out, leaving an equation with only one variable that can be solved.

$$-40y-50y=-1800$$

Add -40y to -50y.

$$-90y=-1800$$

Divide both sides by -90.

$$y=20$$

Substitute 20 for y in x+y=36. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+20=36$$

Subtract 20 from both sides of the equation.

$$x=16$$

The system is now solved.

$$x=16,y=20$$