x = amount of fertilizer 1
y = amount of fertilizer 2
z = amount of fertilizer 3
it helps to make a table for this.
your table would look like this:
amount of fertilizer in kilograms
fertilizer total potassium nitrogen phosphorus
1 x .2x .3x .5x
2 y .1y .2y .7y
3 z --- .3z .7z
4 200 24 50 126
the amounts of potassium, nitrogen, and phosphorous for fertilizer 4 are given to you by weight because you are told the proportions and you know what the total weight is.
potassium = .12 * 200 = 24
nitrogen = .25 * 200 = 50
phosphorus = .63 * 200 = 126
you wind up with 3 equations that need to be solved simultaneously.
note that these equations are from the vertical columns for potassium, nitrogen, and phosphorus in your original table above.
those equations are:
.2x + .1y + .0z = 24 (equation 1)
.3x + .2y + .3z = 50 (equation 2)
.5x + .7y + .7z = 126 (equation 3)
once you solve these simultaneously, you will have the amounts of x,y,z that you need which are the amounts of fertilizer 1, fertilizer 2, fertilizer 3 that you need.
we will eliminate the decimals first so they won't bother us.
multiply all equations by 10 to get:
2x + 1y + 0z = 240 (equation 1)
3x + 2y + 3z = 500 (equation 2)
5x + 7y + 7z = 1260 (equation 3)
first we will take equation 1 and 2 and eliminate the y variable.
we will do that by multiplying equation 1 by 2 and then subtract equation 2 from equation 1 to get equation 4.
4x + 2y + 0z = 480 (equation 1 multiplied by 2)
minus: 3x + 2y + 3z = 500 (equation 2)
equals: 1x - 3z = -20 (equation 4)
next we will take equation 2 and 3 and eliminate the y variable.
we will do that by multiplying equation 2 by 7 and multiplying equation 3 by 2 and then subtracting equation 3 from equation 2.
21x + 14y + 21z = 3500 (equation 2 multiplied by 7)
minus: 10x + 14y + 14z = 2520 (equation 3 multiplied by 2)
equals: 11x + 7z = 980 (equation 5)
next we will take equation 4 and equation 5 and eliminate the x variable.
we will do that by multiplying equation 4 by 11 and then subtracting equation 5 from equation 4
11x - 33z = -220 (equation 4 multiplied by 11)
minus: 11x + 7z = 980 (equation 5)
equals: - 40z = -1200
we solve for z to get:
z = 30
we use this value of z and substitute in equation 5 to get:
11x + 7z = 980 becomes:
11x + 7*(30) = 980 which becomes:
11x + 210 = 980
subtract 210 from both sides to get:
11x = 980-210 = 770
divide both sides by 11 to get:
x = 70
next we use the values of z = 30 and x = 70 and go to equation 3 to get:
5x + 7y + 7z = 1260 becomes:
5*(70) + 7y + 7*30) = 1260 which becomes:
350 + 7y + 210 = 1260 which becomes
7y + 560 = 1260
subtract 560 from both sides to get:
7y = 1260 - 560 = 700
divide both sides by 7 to get:
y = 100
we have values for x,y,z.
they are:
x = 70
y = 100
z = 30
we need to test these values to see if they are good.
our original equations are:
.2x + .1y + .0z = 24 (equation 1)
.3x + .2y + .3z = 50 (equation 2)
.5x + .7y + .7z = 126 (equation 3)
first we check to see if x + y + z = 200.
it does.
that's good because the total of fertilizer 1, 2, 3 should equal 200 and it does.
next we substitute in all 3 equations to get:
.2x + .1y + .0z = 24 (equation 1)
.3x + .2y + .3z = 50 (equation 2)
.5x + .7y + .7z = 126 (equation 3)
becomes:
.2*70 + .1*100 + .0*30 = 24 (equation 1)
.3*70 + .2*100 + .3*30 = 50 (equation 2)
.5*70 + .7*100 + .7*30 = 126 (equation 3)
we simplify to get:
14 + 10 + 0 = 24 (equation 1)
21 + 20 + 9 = 50 (equation 2)
35 + 70 + 21 = 126 (equation 3)
-------------------
total: 70 + 100 + 30 = 200 (sum of equations 1,2,3)
Looks like we have the answer.
he needs:
70 pounds of fertilizer 1
100 pounds of fertilizer 2
30 pounds of fertilizer 3
he will get:
200 pounds of fertilizer 4 that contains:
24 pounds of potassium
50 pounds of nitrogen
126 pounds of phosphorus
