Solve {l}{N_{1}+12=2*(5+N_{2})}{N_1^1-2=3*(N_{2}-3)}

$$\left\{ \begin{array} { l } { N _ { 1 } + 12 = 2 \cdot ( 5 + N _ { 2 } ) } \\ { N _ { 1 } ^ { 1 } - 2 = 3 \cdot ( N _ { 2 } - 3 ) } \end{array} \right.$$

Solve for N_1, N_2

$$N_{1}=8$$

$$N_{2}=5$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 2 by 5+N_{2}.

$$N_{1}+12=10+2N_{2}$$

Subtract 2N_{2} from both sides.

$$N_{1}+12-2N_{2}=10$$

Subtract 12 from both sides.

$$N_{1}-2N_{2}=10-12$$

Subtract 12 from 10 to get -2.

$$N_{1}-2N_{2}=-2$$

Consider the second equation. Calculate N_{1} to the power of 1 and get N_{1}.

$$N_{1}-2=3\left(N_{2}-3\right)$$

Use the distributive property to multiply 3 by N_{2}-3.

$$N_{1}-2=3N_{2}-9$$

Subtract 3N_{2} from both sides.

$$N_{1}-2-3N_{2}=-9$$

Add 2 to both sides.

$$N_{1}-3N_{2}=-9+2$$

Add -9 and 2 to get -7.

$$N_{1}-3N_{2}=-7$$

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.

$$N_{1}-2N_{2}=-2,N_{1}-3N_{2}=-7$$

Choose one of the equations and solve it for N_{1} by isolating N_{1} on the left hand side of the equal sign.

$$N_{1}-2N_{2}=-2$$

Add 2N_{2} to both sides of the equation.

$$N_{1}=2N_{2}-2$$

Substitute -2+2N_{2} for N_{1} in the other equation, N_{1}-3N_{2}=-7.

$$2N_{2}-2-3N_{2}=-7$$

Add 2N_{2} to -3N_{2}.

$$-N_{2}-2=-7$$

Add 2 to both sides of the equation.

$$-N_{2}=-5$$

Divide both sides by -1.

$$N_{2}=5$$

Substitute 5 for N_{2} in N_{1}=2N_{2}-2. Because the resulting equation contains only one variable, you can solve for N_{1} directly.

$$N_{1}=2\times 5-2$$

Multiply 2 times 5.

$$N_{1}=10-2$$

Add -2 to 10.

$$N_{1}=8$$

The system is now solved.

$$N_{1}=8,N_{2}=5$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 2 by 5+N_{2}.

$$N_{1}+12=10+2N_{2}$$

Subtract 2N_{2} from both sides.

$$N_{1}+12-2N_{2}=10$$

Subtract 12 from both sides.

$$N_{1}-2N_{2}=10-12$$

Subtract 12 from 10 to get -2.

$$N_{1}-2N_{2}=-2$$

Consider the second equation. Calculate N_{1} to the power of 1 and get N_{1}.

$$N_{1}-2=3\left(N_{2}-3\right)$$

Use the distributive property to multiply 3 by N_{2}-3.

$$N_{1}-2=3N_{2}-9$$

Subtract 3N_{2} from both sides.

$$N_{1}-2-3N_{2}=-9$$

Add 2 to both sides.

$$N_{1}-3N_{2}=-9+2$$

Add -9 and 2 to get -7.

$$N_{1}-3N_{2}=-7$$

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

$$N_{1}-2N_{2}=-2,N_{1}-3N_{2}=-7$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right)\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=\left(\begin{matrix}-2\\-7\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-2\\1&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right))\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right)\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right))\left(\begin{matrix}-2\\-7\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right))\left(\begin{matrix}-2\\-7\end{matrix}\right)$$

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

$$\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\1&-3\end{matrix}\right))\left(\begin{matrix}-2\\-7\end{matrix}\right)$$

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.

$$\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-\left(-2\right)}&-\frac{-2}{-3-\left(-2\right)}\\-\frac{1}{-3-\left(-2\right)}&\frac{1}{-3-\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}-2\\-7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=\left(\begin{matrix}3&-2\\1&-1\end{matrix}\right)\left(\begin{matrix}-2\\-7\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=\left(\begin{matrix}3\left(-2\right)-2\left(-7\right)\\-2-\left(-7\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}N_{1}\\N_{2}\end{matrix}\right)=\left(\begin{matrix}8\\5\end{matrix}\right)$$

Extract the matrix elements N_{1} and N_{2}.

$$N_{1}=8,N_{2}=5$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 2 by 5+N_{2}.

$$N_{1}+12=10+2N_{2}$$

Subtract 2N_{2} from both sides.

$$N_{1}+12-2N_{2}=10$$

Subtract 12 from both sides.

$$N_{1}-2N_{2}=10-12$$

Subtract 12 from 10 to get -2.

$$N_{1}-2N_{2}=-2$$

Consider the second equation. Calculate N_{1} to the power of 1 and get N_{1}.

$$N_{1}-2=3\left(N_{2}-3\right)$$

Use the distributive property to multiply 3 by N_{2}-3.

$$N_{1}-2=3N_{2}-9$$

Subtract 3N_{2} from both sides.

$$N_{1}-2-3N_{2}=-9$$

Add 2 to both sides.

$$N_{1}-3N_{2}=-9+2$$

Add -9 and 2 to get -7.

$$N_{1}-3N_{2}=-7$$

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.

$$N_{1}-2N_{2}=-2,N_{1}-3N_{2}=-7$$

Subtract N_{1}-3N_{2}=-7 from N_{1}-2N_{2}=-2 by subtracting like terms on each side of the equal sign.

$$N_{1}-N_{1}-2N_{2}+3N_{2}=-2+7$$

Add N_{1} to -N_{1}. Terms N_{1} and -N_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-2N_{2}+3N_{2}=-2+7$$

Add -2N_{2} to 3N_{2}.

$$N_{2}=-2+7$$

Add -2 to 7.

$$N_{2}=5$$

Substitute 5 for N_{2} in N_{1}-3N_{2}=-7. Because the resulting equation contains only one variable, you can solve for N_{1} directly.

$$N_{1}-3\times 5=-7$$

Multiply -3 times 5.

$$N_{1}-15=-7$$

Add 15 to both sides of the equation.

$$N_{1}=8$$

The system is now solved.

$$N_{1}=8,N_{2}=5$$