Question

How would I solve this equation using the linear combination methodn=m+1m=2n+2

Answer 1

We must solve, using the linear combination method, the following system of equations:

`\begin{cases}n=m+1 \\ m=2n+1\end{cases}`

(1) We reorder the variables in the equations such that the variables are ordered in columns

We rewrite the first equation in the following way:

`\begin{gathered} n=m+1, \\ n-m=m+1-m, \\ n-m=1. \end{gathered}`

We rewrite the second equation in the following way:

`\begin{gathered} m=2n+1, \\ m-2n=1, \\ -2n+m=1. \end{gathered}`

So, our system of equation is equivalent to:

`\begin{cases}n-m=1 \\ -2n+m=1\end{cases}`

2) We solve for n

Summing both equations, we get:

`\begin{gathered} (n-2n)+(m-m)=(1+1) \\ -n=2, \\ n=-2. \end{gathered}`

NOTE: When we sum the equations:

0. we sum the left side of eq. 1 with the left side of eq. 2,

,

1. we sum the right side of eq. 1 with the right side of eq. 2,

3) We solve for m

We multiply the first equation by 2:

`\begin{cases}2n-2m=2 \\ -2n+m=1\end{cases}`

Summing both equations, we get:

`\begin{gathered} (2n-2n)+(-2m+m)=(2+1), \\ -m=3, \\ m=-3. \end{gathered}`

Answer

The solution of the system of equations is:

`\begin{gathered} n=-2 \\ m=-3 \end{gathered}`