Mrs. Alderidge operates a soybean farm outside of Grinnell, Iowa. To keep costs down, she buys many products in bulk and transfers them to s

Here's another way to do this with only one variable.

Call the amount of the 63% solution = "x"...then the amount of the 43% solution is just "40-x"..we have

.63x + .43(40-x)  = .53(40)     multiply through by 100

63x + 43(40-x) = 53(40)

63x + 1720 - 43x = 2120      simplify

20x  = 400       divide through by 20

x= 20 L

Let X be the amount of 63% soln used and let Y be the amount of 43% solution used.

$$\begin{array}{rlll} X+Y&=&40\qquad \qquad & (1)\\ 0.63X+0.43Y&=&0.53(X+Y)& (2)\\ 0.20X+(0.43X+0.43Y)&=&0.53*40\qquad $Error fixed - thanks Alan$\\ 0.20X+0.43(X+Y)&=&21.2\\ 0.20X+0.43*40&=&21.2\\ 0.20X+17.2&=&21.2\\ 0.20X&=&4\\ X&=&4/0.20\\ X&=&20\\ \end{array}$$

so it seems that 20L of the 63% solution was needed.  

Maybe you were supposed to ESTIMATE the answer

A solution of 63% fertilizer is to be mixed with a solution of 43% fertilizer to form 40 liters of a 53% solution. How much of the 63% solution must she use?

Choices     20L, 34L, 15L, 32L

The 63% solution is too strong by 10% and the 43% is too week by 10% so if i had to guess I would say 50% of each is needed.  That is 20L each.       

Again: This answer has  been edited.

Melody, the first term in the third line of your derivation should be 0.2X not 0.23X.  

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Thanks Alan,

Much appreciated :)