Question

The first order was for 13 bushes and 4 trees, and had a total of $432. The second order was for 6 bushes and 2 trees, and had a total of $232. what was the cost of one bush?

Answer 1

DISCLAIMER: One of the solutions of this system is negative, which seems strange. Make sure there are non typos in the question!

Let b be the price of a bush and t the price of a tree. The two orders translate to the following linear system:

$\begin{cases}13b+4t = 432\\6b+2t=232 \end{cases}$

Multiply the second equation by 2:

$\begin{cases}13b+4t = 432\\12b+4t=464 \end{cases}$

Subtract the second equation from the first:

$(13b+4t)-(12b+4t) = 432-464 \iff b = -32$

Substitute this value in the first equation (for example) to compute the value of t:

$13b+4t = 432 \iff 13\cdot(-32) + 4t = 432 \iff 4t = 432+13\cdot32 = 848 \iff t = 212$