Solve {l}{1/2(x-80+120)=600-N_{1}}{1/2(x-40+120)=600-12/13N_{1}}

$$\left. \begin{array} { l } { \frac { 1 } { 2 } ( x - 80 + 120 ) = 600 - N _ { 1 } } \\ { \frac { 1 } { 2 } ( x - 40 + 120 ) = 600 - \frac { 12 } { 13 } N _ { 1 } } \end{array} \right.$$

Solve for x, N_1

$$x=640$$

$$N_{1}=260$$

Steps Using Substitution

Consider the first equation. Add -80 and 120 to get 40.

$$\frac{1}{2}\left(x+40\right)=600-N_{1}$$

Use the distributive property to multiply \frac{1}{2} by x+40.

$$\frac{1}{2}x+20=600-N_{1}$$

Add N_{1} to both sides.

$$\frac{1}{2}x+20+N_{1}=600$$

Subtract 20 from both sides.

$$\frac{1}{2}x+N_{1}=600-20$$

Subtract 20 from 600 to get 580.

$$\frac{1}{2}x+N_{1}=580$$

Consider the second equation. Add -40 and 120 to get 80.

$$\frac{1}{2}\left(x+80\right)=600-\frac{12}{13}N_{1}$$

Use the distributive property to multiply \frac{1}{2} by x+80.

$$\frac{1}{2}x+40=600-\frac{12}{13}N_{1}$$

Add \frac{12}{13}N_{1} to both sides.

$$\frac{1}{2}x+40+\frac{12}{13}N_{1}=600$$

Subtract 40 from both sides.

$$\frac{1}{2}x+\frac{12}{13}N_{1}=600-40$$

Subtract 40 from 600 to get 560.

$$\frac{1}{2}x+\frac{12}{13}N_{1}=560$$

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.

$$\frac{1}{2}x+N_{1}=580,\frac{1}{2}x+\frac{12}{13}N_{1}=560$$

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

$$\frac{1}{2}x+N_{1}=580$$

Subtract N_{1} from both sides of the equation.

$$\frac{1}{2}x=-N_{1}+580$$

Multiply both sides by 2.

$$x=2\left(-N_{1}+580\right)$$

Multiply 2 times -N_{1}+580.

$$x=-2N_{1}+1160$$

Substitute -2N_{1}+1160 for x in the other equation, \frac{1}{2}x+\frac{12}{13}N_{1}=560.

$$\frac{1}{2}\left(-2N_{1}+1160\right)+\frac{12}{13}N_{1}=560$$

Multiply \frac{1}{2} times -2N_{1}+1160.

$$-N_{1}+580+\frac{12}{13}N_{1}=560$$

Add -N_{1} to \frac{12N_{1}}{13}.

$$-\frac{1}{13}N_{1}+580=560$$

Subtract 580 from both sides of the equation.

$$-\frac{1}{13}N_{1}=-20$$

Multiply both sides by -13.

$$N_{1}=260$$

Substitute 260 for N_{1} in x=-2N_{1}+1160. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times 260+1160$$

Multiply -2 times 260.

$$x=-520+1160$$

Add 1160 to -520.

$$x=640$$

The system is now solved.

$$x=640,N_{1}=260$$

Steps Using Matrices

Consider the first equation. Add -80 and 120 to get 40.

$$\frac{1}{2}\left(x+40\right)=600-N_{1}$$

Use the distributive property to multiply \frac{1}{2} by x+40.

$$\frac{1}{2}x+20=600-N_{1}$$

Add N_{1} to both sides.

$$\frac{1}{2}x+20+N_{1}=600$$

Subtract 20 from both sides.

$$\frac{1}{2}x+N_{1}=600-20$$

Subtract 20 from 600 to get 580.

$$\frac{1}{2}x+N_{1}=580$$

Consider the second equation. Add -40 and 120 to get 80.

$$\frac{1}{2}\left(x+80\right)=600-\frac{12}{13}N_{1}$$

Use the distributive property to multiply \frac{1}{2} by x+80.

$$\frac{1}{2}x+40=600-\frac{12}{13}N_{1}$$

Add \frac{12}{13}N_{1} to both sides.

$$\frac{1}{2}x+40+\frac{12}{13}N_{1}=600$$

Subtract 40 from both sides.

$$\frac{1}{2}x+\frac{12}{13}N_{1}=600-40$$

Subtract 40 from 600 to get 560.

$$\frac{1}{2}x+\frac{12}{13}N_{1}=560$$

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

$$\frac{1}{2}x+N_{1}=580,\frac{1}{2}x+\frac{12}{13}N_{1}=560$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right)\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=\left(\begin{matrix}580\\560\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right))\left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right)\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right))\left(\begin{matrix}580\\560\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right))\left(\begin{matrix}580\\560\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&1\\\frac{1}{2}&\frac{12}{13}\end{matrix}\right))\left(\begin{matrix}580\\560\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{12}{13}}{\frac{1}{2}\times \frac{12}{13}-\frac{1}{2}}&-\frac{1}{\frac{1}{2}\times \frac{12}{13}-\frac{1}{2}}\\-\frac{\frac{1}{2}}{\frac{1}{2}\times \frac{12}{13}-\frac{1}{2}}&\frac{\frac{1}{2}}{\frac{1}{2}\times \frac{12}{13}-\frac{1}{2}}\end{matrix}\right)\left(\begin{matrix}580\\560\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=\left(\begin{matrix}-24&26\\13&-13\end{matrix}\right)\left(\begin{matrix}580\\560\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=\left(\begin{matrix}-24\times 580+26\times 560\\13\times 580-13\times 560\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\N_{1}\end{matrix}\right)=\left(\begin{matrix}640\\260\end{matrix}\right)$$

Extract the matrix elements x and N_{1}.

$$x=640,N_{1}=260$$

Steps Using Elimination

Consider the first equation. Add -80 and 120 to get 40.

$$\frac{1}{2}\left(x+40\right)=600-N_{1}$$

Use the distributive property to multiply \frac{1}{2} by x+40.

$$\frac{1}{2}x+20=600-N_{1}$$

Add N_{1} to both sides.

$$\frac{1}{2}x+20+N_{1}=600$$

Subtract 20 from both sides.

$$\frac{1}{2}x+N_{1}=600-20$$

Subtract 20 from 600 to get 580.

$$\frac{1}{2}x+N_{1}=580$$

Consider the second equation. Add -40 and 120 to get 80.

$$\frac{1}{2}\left(x+80\right)=600-\frac{12}{13}N_{1}$$

Use the distributive property to multiply \frac{1}{2} by x+80.

$$\frac{1}{2}x+40=600-\frac{12}{13}N_{1}$$

Add \frac{12}{13}N_{1} to both sides.

$$\frac{1}{2}x+40+\frac{12}{13}N_{1}=600$$

Subtract 40 from both sides.

$$\frac{1}{2}x+\frac{12}{13}N_{1}=600-40$$

Subtract 40 from 600 to get 560.

$$\frac{1}{2}x+\frac{12}{13}N_{1}=560$$

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.

$$\frac{1}{2}x+N_{1}=580,\frac{1}{2}x+\frac{12}{13}N_{1}=560$$

Subtract \frac{1}{2}x+\frac{12}{13}N_{1}=560 from \frac{1}{2}x+N_{1}=580 by subtracting like terms on each side of the equal sign.

$$\frac{1}{2}x-\frac{1}{2}x+N_{1}-\frac{12}{13}N_{1}=580-560$$

Add \frac{x}{2} to -\frac{x}{2}. Terms \frac{x}{2} and -\frac{x}{2} cancel out, leaving an equation with only one variable that can be solved.

$$N_{1}-\frac{12}{13}N_{1}=580-560$$

Add N_{1} to -\frac{12N_{1}}{13}.

$$\frac{1}{13}N_{1}=580-560$$

Add 580 to -560.

$$\frac{1}{13}N_{1}=20$$

Multiply both sides by 13.

$$N_{1}=260$$

Substitute 260 for N_{1} in \frac{1}{2}x+\frac{12}{13}N_{1}=560. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{1}{2}x+\frac{12}{13}\times 260=560$$

Multiply \frac{12}{13} times 260.

$$\frac{1}{2}x+240=560$$

Subtract 240 from both sides of the equation.

$$\frac{1}{2}x=320$$

Multiply both sides by 2.

$$x=640$$

The system is now solved.

$$x=640,N_{1}=260$$