Hiya. I'm going to rewrite the second equation.
By subtracting 5x by both sides, I'll be able to have y by itself:
-y=-5x+13
I'm going to then divide both sides by -1 to get:
y=5x-13.
Then, I'm going to plug that equation into the first equation
3x+2(5x-13)=39
Factor:
3x+10x-26=39
Combine like terms:
13x=65
Divide both sides by 13 to get x by itself to get x=5
Plug this back into the equation of y=5x-13
y=5(5)-13 to get y=12
I don’t know why the word “million” is there, but the incremental cost per unit of the secon hundred units is $30 per unit.
Im pretty sure you are i have done this question before and im pretty sure thats what i put and i got it right. plus just go with your gut
a = amount invested at 5%
b = amount invested at 2%
now, we know the total invested by Mr Wilson was 20000, so whatever "a" and "b" might be, we know that a + b = 20000.
![a + b = 20000\implies b = 20000-a \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{5\% of a}}{\left( \cfrac{5}{100} \right)a}\implies 0.05a~\hfill \stackrel{\textit{2\% of b}}{\left( \cfrac{2}{100} \right)a}\implies \begin{array}{llll} 0.02b\\\\ \stackrel{substituting}{0.02(20000-a)} \end{array}](https://tex.z-dn.net/?f=a%20%2B%20b%20%3D%2020000%5Cimplies%20b%20%3D%2020000-a%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B5%5C%25%20of%20a%7D%7D%7B%5Cleft%28%20%5Ccfrac%7B5%7D%7B100%7D%20%5Cright%29a%7D%5Cimplies%200.05a~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B2%5C%25%20of%20b%7D%7D%7B%5Cleft%28%20%5Ccfrac%7B2%7D%7B100%7D%20%5Cright%29a%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%200.02b%5C%5C%5C%5C%20%5Cstackrel%7Bsubstituting%7D%7B0.02%2820000-a%29%7D%20%5Cend%7Barray%7D)
now, we know the total of earned interest is 550 bucks, so then
25/12.5 = 40/y
cross multiply
(25)(y) = (12.5)(40)
25y = 500
y = 500/25
y = 20 <====
