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>
<em />
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.
Answer:
option H.
Line A increase more rapidly than B
hope it helps
Answer:
We have;
7 pencil boxes and 5 eraser boxes
Step-by-step explanation:
To find the smallest number of boxes that can be bought, what we simply do is to find the lowest common multiple of the content of each of the boxes
hence, we are going to find the lowest common multiple of 10 and 14
The lowest common multiple of 10 and 14 is 70
So the smallest number of pencil boxes that can be bought is 70/10 = 7
The smallest number of eraser boxes would be 70/14 = 5
Now, let's say there are "b" boys and "g" gals... ok... well, we know there are 116 more boys than gals..... so, if there are "g" gals then there must be "g + 116" boys.
what's the total class? well is g + b or g + ( g + 116).
