Elimination method:
4m = n + 7
3m + 4n + 9 = 0
<em>First, let's get the equations in the same form.</em>
4m - n - 7 = 0
3m + 4n + 9 = 0
<em>Now let's make multiply the first equation by 4 so we can eliminate n.</em>
16m - 4n - 28 = 0
+3m + 4n + 9 = 0
<em>Now we can add the equations.</em>
16m + 3m - 4n + 4n - 28 + 9 = 0
19m + 0n - 19 = 0
19m - 19 = 0
19m = 19
<em>m = 1</em>
<em>Now we put m back into one (or both) of the original equations.</em>
4(1) = n + 7
4 = n + 7
<em>n = -3</em>
<em>If you plug m into the other equation, you get the same result.</em>
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Substitution method:
4m = n + 7
3m + 4n + 9 = 0
<em>With this method, we plug one of the equations into the other one. I'm going to use m in the second equation as a substitute for m in the second equation.</em>
3m + 4n + 9 = 0
3m = -4n - 9
m = (-4/3)n - 3
<em>Now I can substitute the right side into the first equation like so:</em>
4[(-4/3)n - 3] = n + 7
(-16n)/3 - 12 = n + 7
(-16n)/3 = n + 19
-16n = 3(n + 19)
-16n = 3n + 57
0 = 16n + 3n + 57
0 = 19n + 57
0 = 19n/19 + 57/19
0 = n + 3
<em>-3 = n</em>
<em>And then we put that back into one of the original equations.</em>
4m = n + 7
4m = -3 + 7
4m = 4
<em>m = 1</em>
Hopefully you learned something from this.
All three clocks will show the same time in 1 day.
The answer i think it is, is C
