For the treatment

For the treatment of seedlings in a forest nursery there are available 4 types of preparations P1, P2, P3, P4, each of which contains 3 active substances L1, L2, L3. Their amounts are given by the table (number of units/kg). It is necessary to mix 20 kg of a mixture so that the produced amount contains 34 units of substance L1, 41 units of substance L2, and 32 units of substance L3. Find out how many kg of preparations P1, P2, P3 and P4 need to be used to mix a mixture with the required properties.


P1 P2 P3 P4
L1 2 3 1 1
L2 3 2 2 1
L3 1 2 2 2

Final Answer:

a =  8
b =  3
c =  2
d =  7

Step-by-step explanation:


2a+3b+c+d = 34
3a+2b+2c+d = 41
a+2b+2c+2d = 32
a+b+c+d = 20

2·a+3·b+c+d = 34
3·a+2·b+2·c+d = 41
a+2·b+2·c+2·d = 32
a+b+c+d = 20

2a+3b+c+d = 34
3a+2b+2c+d = 41
a+2b+2c+2d = 32
a+b+c+d = 20

Pivot: Row 1 ↔ Row 2
3a+2b+2c+d = 41
2a+3b+c+d = 34
a+2b+2c+2d = 32
a+b+c+d = 20

Row 2 - 2/3 · Row 1 → Row 2
3a+2b+2c+d = 41
1.6667b-0.3333c+0.3333d = 6.6667
a+2b+2c+2d = 32
a+b+c+d = 20

Row 3 - 1/3 · Row 1 → Row 3
3a+2b+2c+d = 41
1.6667b-0.3333c+0.3333d = 6.6667
1.3333b+1.3333c+1.6667d = 18.3333
a+b+c+d = 20

Row 4 - 1/3 · Row 1 → Row 4
3a+2b+2c+d = 41
1.6667b-0.3333c+0.3333d = 6.6667
1.3333b+1.3333c+1.6667d = 18.3333
0.3333b+0.3333c+0.6667d = 6.3333

Row 3 - 1.33333333/1.66666667 · Row 2 → Row 3
3a+2b+2c+d = 41
1.6667b-0.3333c+0.3333d = 6.6667
1.6c+1.4d = 13
0.3333b+0.3333c+0.6667d = 6.3333

Row 4 - 0.33333333/1.66666667 · Row 2 → Row 4
3a+2b+2c+d = 41
1.6667b-0.3333c+0.3333d = 6.6667
1.6c+1.4d = 13
0.4c+0.6d = 5

Row 4 - 0.4/1.6 · Row 3 → Row 4
3a+2b+2c+d = 41
1.6667b-0.3333c+0.3333d = 6.6667
1.6c+1.4d = 13
0.25d = 1.75


d = 1.75/0.25 = 7
c = 13-1.4d/1.6 = 13-1.4 · 7/1.6 = 2
b = 6.66666667+0.33333333333333c-0.33333333333333d/1.66666667 = 6.66666667+0.33333333 · 2-0.33333333 · 7/1.66666667 = 3
a = 41-2b-2c-d/3 = 41-2 · 3-2 · 2-7/3 = 8

a = 8
b = 3
c = 2
d = 7

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