M+4n=8
m=n-2
we see that m=n-2
so we can subsitute 'n-2' for m in the first equation
(n-2)+4n=8
n-2+4n=8
5n-2=8
add 2 to both sides
5n=10
divide by 5
n=2
subsitute
n=2
m=n-2
m=2-2
m=0
m=0
n=2
the answer is c. 0,2
The box has volume
11 in × 3 in × <em>x</em> in = 33<em>x</em> in³
and surface area
2 (11 in × 3 in + 11 in × <em>x</em> in + 3 in × <em>x</em> in) = (66 + 28<em>x</em>) in²
The volume and area have the same value if
33<em>x</em> = 66 + 28<em>x</em>
Solve this equation for <em>x</em> :
33<em>x</em> - 28<em>x</em> = 66
5<em>x</em> = 66
<em>x</em> = 66/5 = 13.2
I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
C. Because it equals to the degrees to the 3.7 radians
