Question

Solve the system, using substitution. What is the n coordinate of the solution?
2m - 3n = 6 and m = 3 - 2n


A. 2
B. 0
C. -4
D. 2
E. no solution

Answer 1

\[2(3-2n)-3n=6\Rightarrow 6-4n-3n=6\Rightarrow-7n=6-6\Rightarrow-7n=0\Rightarrow n=0\]

Answer 2

Thanks again John! I really appreciate you helping me :)

Answer 3

So we have, two equations. Usually, so include a general solution: with two linear equations in the form of \(y=mx+b\), one can isolate like so:
\[\eqalign{
&1.\phantom{space}y=s_1x+b_1 \\
&2.\phantom{space}y=s_2x+b_2 \\
&\\
&s_1x+b_1=s_2x+b_2 \\
&(s_1-s_2)x=b_2-b_1 \\
&\\
&x=\frac{b_2-b_1}{s_1-s_2} \\
}\]

So let us rearrange both equations into this form:
\[\eqalign{
&2m-3n=6 \\
&2m=6+3n \\
&m=\frac{3n+6}{2} \\
&m=(1.5)n+3 \\
}
\phantom{space}\eqalign{
&m=3-2n \\
&m=(-2)n+3
}\]
Therefore, we can derive the following values:
\[\eqalign{
&s_1=1.5 \\
&b_1=3 \\
&s_2=-2 \\
&b_2= 3 \\
}\]
So we can plug into equation:
\[n=\frac{b_2-b_1}{s_1-s_2}=\frac{3-3}{1.5-(-2)}=\frac{0}{1.5+2}=\frac{0}{3.5}=0\]

Therfore, \(n=0\)