SOLUTION: Solve using substitution/Addition By weight a 10-10-10 fertilizer contains 10.0% nitrogen, 10.0% phosphorus, and 10.0% potash. A 12-0-6 fertilizer contains 12.0% nitrogen, no

Question 1145000: Solve using substitution/Addition
By weight a 10-10-10 fertilizer contains 10.0% nitrogen, 10.0% phosphorus, and 10.0% potash. A 12-0-6
fertilizer contains 12.0% nitrogen, no phosphorous, and 6.00% potash. A landscaper has three types of
fertilizer in stock. One is 10-12-15, a second is 10-0-5, and a third is 30-6-15. How much of each must
be mixed in order to get 400̃ lb of fertilizer that is 16-3-9?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your simultaneous equations to solve become:

.10x + .10y + .30z = 64
.12x + 0y + .06z = 12
.15x + .05y + .15z = 36

the first equation tells you the total number of pounds of nitrogen you want.
the second equation tells you the total number of pounds of phosphorus you want.
the third equation tells you the total number of pounds of potash you want.

these are all based on the total number of pounds of fertilizer being equal to 400 pounds.

x is the number of pounds of fertilizer that is 10-12-15.
y is the number of pounds of fertilizer that is 10-0-5.
z is the number of pounds of fertilizer that is 30-6-15.

the total number of pounds of fertilizer that you want is 400.
that is composed of 16% nitrogen = 64 pounds of nitrogen.
that is composed of 3% phosphorus = 12 pounds of phosphorus.
that is composed of 9% potash= 36 pounds of potash.

the 3 equations you want to solve simultaneously are:

.10x + .10y + .30z = 64
.12x + 0y + .06z = 12
.15x + .05y + .15z = 36

take the first 2 equations and eliminate the x variable by multiplying both sides of the first equation by 1.2 and then subtracting the second equation from the first to get:
.12y + .30z = 64.8

take the second 2 equations and eliminate the x variable by multiplying both sides of second equation by 1.25 and then subtracting the second equation from the third to get:
.05y + .075z = 21

you now have two equations in y and z.

they are:

.12y + .30z = 64.8
.05y + .075z = 21

eliminate y by multiplying both sides of the second equation by 2.4 and leaving the first equation as is to get:

.12y + .30z = 64.8
.12y + .18z = 50.4

subtract the second equation from the first to ge:

.12z = 14.4
solve for z to get:
z = 14.4 /.12 = 120

replace z in the second equation to get:
.12y + .30z = 64.8 becomes:
.12y + .30 * 120 = 64.8 which becomes:
.12y + 36 = 64.8
subtract 36 from both sides of this equation to get:
.12y = 28.8
solve for y to get:
y = 28.8 / .12 = 240

you now have:
z = 120
y = 240

go back to the first original equation and replace z with 120 and y with 240 to get:L
.10x + .10y + .30z = 64 becomes:
.10x + .10 * 240 + .30 * 120 = 64 which becomes:
.10x + 24 + 36 = 64
combine like terms to get:
.10x + 60 = 64
subtract 60 from both sides of this equation to get:
.10x = 4
solve for x to get:
x = 4 / .10 = 40

your solution is that:
x = 40
y = 240
z = 120

you will need:

40 pounds of fertilizer that is 10-12-15.
240 pounds of fertilizer that is 10-0-5.
120 pounds of fertilizer that is 30-6-15.

total pounds of fertilizer is equal to 40 + 240 + 120 = 400
total pounds of nitrogen is .10 * 40 + .10 * 240 + .30 * 120 = 64
total pounds of phosphorus is .12 * 40 + 0 * 240 + .06 * 120 = 12
total pounds of potash is .15 * 40 + .05 * 240 + .15 * 120 = 36

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