Solving for 4 variables-algebra

Question

I encountered a problem with 4 variables and I was wondering if anyone knows how to solve this: This is what is known:

$$ \left\lbrace \begin{align} a+b &= 1800 \\ c+d &= 12 \\ a/c &= 100 \\ b/d &= 250 \\ (a+b)/(c+d) &= 150 \end{align} \right.$$

Below is a screenshot from a spreadsheet. The red numbers are the 4 unknowns that I'm trying to figure out how to solve for (I happen to know them, but would love to understand how to solve for them when I do not know them). screenshot Any help would be greatly appreciated! Thank you!

Answer 1

$$a/c = 100 \implies a = 100c \\ b/d = 250 \implies b = 250d$$

Substitute this into $a+b = 1800$ to get $100c+250d = 1800$. Divide by $50$ to get $2c+5d = 36$. From $c+d = 12$, we have $c = 12-d$, so $2(12-d)+5d = 36$.

$3d = 12 \iff d=4$, so $c = 12-4=8$. Hence $(a,b,c,d) = (800,1000,8,4)$

Answer 2

Linear algebra will do ... we have \begin{eqnarray*} a+b=1800 \\ c+d=12 \end{eqnarray*} and $a=100c$, $b=250d$ sub these into the first equation & cancel a factor of $50$. (Note that the fifth equation is satisfied by virtue of the first two.) \begin{eqnarray*} 2c+5d=36 \\ c+d=12 \end{eqnarray*} Should be a doddle from here ?

Answer 3

$b=1800-a$

$d=12-c$

$\cfrac ac=100$

$a=100c$

$\cfrac {1800-a}{12-c}=250$

$\cfrac {2.5a}c=250$

$\cfrac {1800-a}{12-c}=\cfrac {2.5a}c$

$\cfrac {1800-100c}{12-c}=\cfrac {2.5(100c)}c$

$1800c-10c^2=250c(12-c)$

Which gives you the value of $c$