Solve {l}{x+y=850}{3*16x=2*10y(2)}

$$\left\{ \begin{array} { l } { x + y = 850 } \\ { 3 \times 16 x = 2 \times 10 y ( 2 ) } \end{array} \right.$$

Solve for x, y

$$x = \frac{4250}{11} = 386\frac{4}{11} \approx 386.363636364$$

$$y = \frac{5100}{11} = 463\frac{7}{11} \approx 463.636363636$$

Steps Using Substitution

Consider the second equation. Subtract 2\times 10y\times 2 from both sides.

$$3\times 16x-2\times 10y\times 2=0$$

Multiply -1 and 2 to get -2.

$$3\times 16x-2\left(2\times 10\right)y=0$$

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.

$$x+y=850,48x-40y=0$$

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

$$x+y=850$$

Subtract y from both sides of the equation.

$$x=-y+850$$

Substitute -y+850 for x in the other equation, 48x-40y=0.

$$48\left(-y+850\right)-40y=0$$

Multiply 48 times -y+850.

$$-48y+40800-40y=0$$

Add -48y to -40y.

$$-88y+40800=0$$

Subtract 40800 from both sides of the equation.

$$-88y=-40800$$

Divide both sides by -88.

$$y=\frac{5100}{11}$$

Substitute \frac{5100}{11} for y in x=-y+850. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5100}{11}+850$$

Add 850 to -\frac{5100}{11}.

$$x=\frac{4250}{11}$$

The system is now solved.

$$x=\frac{4250}{11},y=\frac{5100}{11}$$

Steps Using Matrices

Consider the second equation. Subtract 2\times 10y\times 2 from both sides.

$$3\times 16x-2\times 10y\times 2=0$$

Multiply -1 and 2 to get -2.

$$3\times 16x-2\left(2\times 10\right)y=0$$

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

$$x+y=850,48x-40y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\48&-40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}850\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\48&-40\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\48&-40\end{matrix}\right))\left(\begin{matrix}1&1\\48&-40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\48&-40\end{matrix}\right))\left(\begin{matrix}850\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\48&-40\end{matrix}\right))\left(\begin{matrix}850\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\48&-40\end{matrix}\right))\left(\begin{matrix}850\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{40}{-40-48}&-\frac{1}{-40-48}\\-\frac{48}{-40-48}&\frac{1}{-40-48}\end{matrix}\right)\left(\begin{matrix}850\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{11}&\frac{1}{88}\\\frac{6}{11}&-\frac{1}{88}\end{matrix}\right)\left(\begin{matrix}850\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{11}\times 850\\\frac{6}{11}\times 850\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4250}{11}\\\frac{5100}{11}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{4250}{11},y=\frac{5100}{11}$$

Steps Using Elimination

Consider the second equation. Subtract 2\times 10y\times 2 from both sides.

$$3\times 16x-2\times 10y\times 2=0$$

Multiply -1 and 2 to get -2.

$$3\times 16x-2\left(2\times 10\right)y=0$$

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.

$$x+y=850,48x-40y=0$$

To make x and 48x equal, multiply all terms on each side of the first equation by 48 and all terms on each side of the second by 1.

$$48x+48y=48\times 850,48x-40y=0$$

Simplify.

$$48x+48y=40800,48x-40y=0$$

Subtract 48x-40y=0 from 48x+48y=40800 by subtracting like terms on each side of the equal sign.

$$48x-48x+48y+40y=40800$$

Add 48x to -48x. Terms 48x and -48x cancel out, leaving an equation with only one variable that can be solved.

$$48y+40y=40800$$

Add 48y to 40y.

$$88y=40800$$

Divide both sides by 88.

$$y=\frac{5100}{11}$$

Substitute \frac{5100}{11} for y in 48x-40y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$48x-40\times \frac{5100}{11}=0$$

Multiply -40 times \frac{5100}{11}.

$$48x-\frac{204000}{11}=0$$

Add \frac{204000}{11} to both sides of the equation.

$$48x=\frac{204000}{11}$$

Divide both sides by 48.

$$x=\frac{4250}{11}$$

The system is now solved.

$$x=\frac{4250}{11},y=\frac{5100}{11}$$