Answer:
a) p for plain wrapping, h for holiday wrapping
b) ![\left \{ {{7p+4h=68} \atop {6p+2h=44}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B7p%2B4h%3D68%7D%20%5Catop%20%7B6p%2B2h%3D44%7D%7D%20%5Cright.)
c) Plain wrapping paper cost $4/roll, and holiday wrapping paper costs $10/roll.
Step-by-step explanation:
For part a, I've decided to with p for plain and h for holiday. Part b is somewhat simple;
is the system we'll be using.
Now to do the actual work with part c.
I'll start by solving for p in 7p + 4h = 68.
7p + 4h = 68
<em>Step 1: Subtract 4h from both sides.</em>
7p = -4h + 68
<em>Step 2: Divide both sides by 7/Multiply both sides by 1/7 (Either can be done)</em>
p = 1/7(-4h + 68)
<em>Step 3: Multiply 1/7 by -4h + 68</em>
p = -4h/7 + 68/7
Now substitute -4h + 68/7 for p in the other equation (6p + 2h = 44).
6(-4h/7 + 68/7) + 2h = 44
<em>Step 4: Multiply 6 by -4h/7 + 68/7.</em>
-24h/7 + 408/7 + 2h = 44
<em>Step 5: Add -24h/7 to 2h.</em>
-10h/7 + 408/7 = 44
<em>Step 6: Subtract 408/7 from both sides of the equation.</em>
-10h/7 = -100/7
<em>Step 7: Multiply both sides by -7/10, the reciprocal of -10/7.</em>
h = 10
Now we know the holiday wrapping paper is $10/roll. Substitute 10 for h in p = -4h/7 + 68/7
p = 4(10)/7 + 68/7
<em>Step 8: Multiply -4 by 10 to get -40.</em>
p = -40/7 + 68/7
<em>Step 9 (This is optional): Make what we have from Step 8 a single fraction.</em>
p = (-40 + 68)/7
<em>Steps 10 and 11: Add -40 and 68 to get 28, then divide that by 7 to get 4.</em>
p = 4
And now we know the plain wrapping paper is $4/roll.
So, the complete sentence is: Plain wrapping paper cost $4/roll, and holiday wrapping paper costs $10/roll.