I'm not sure of how to solve word problems. The question is,"One lawn fertilizer is 24% nitrogen and another is 12% nitrogen. How much of fertilizer should be mixed to obtain 100 kg of fertilizer that is 21% nitrogen?"

see the other post.

Let m1 and m2 be the masses of fertilizer at 24% and 21% respectively. Then write:

Eq. 1: m1 + m2 = 100
Eq : (.24m1 + .21 m2)/(m1 + m2)= .21

The second equation is called a weighted average. Solving simultaneously yields:

m1 = 75 kg
m2 = 25 kg

Checking:

(.24×75 + .21× 25)/100 = .18 + .03 = .21
QED

Let x be mass of 24% mixture. Let y be mass of 12% mixture.

Equation would be, x+y=100
0.24x+0.12y= (0.21)(100)

To solve this word problem, you need to use the concept of mixture or combination. Here's how you can approach it:

Step 1: Identify the given information:
- One fertilizer is 24% nitrogen.
- Another fertilizer is 12% nitrogen.
- You want to obtain 100 kg of fertilizer that is 21% nitrogen.

Step 2: Set up the equation:
Let's assume x kg of the 24% nitrogen fertilizer is mixed with (100 - x) kg of the 12% nitrogen fertilizer to obtain 100 kg of the desired 21% nitrogen fertilizer.

Step 3: Express the nitrogen content mathematically:
The nitrogen content of the 24% nitrogen fertilizer can be expressed as 0.24x, since it contains 24% or 0.24 of nitrogen per kg.
Similarly, the nitrogen content of the 12% nitrogen fertilizer can be expressed as 0.12(100 - x), as it contains 12% or 0.12 of nitrogen per kg.

The nitrogen content of the resulting 21% nitrogen fertilizer can be expressed as 0.21(100), as it contains 21% or 0.21 of nitrogen per kg.

Step 4: Set up the equation based on the nitrogen content:
0.24x + 0.12(100 - x) = 0.21(100)

Step 5: Solve the equation:
Solve the equation using basic algebraic methods. Here's the step-by-step solution:
0.24x + 0.12(100 - x) = 0.21(100)
0.24x + 12 - 0.12x = 21
0.12x = 21 - 12
0.12x = 9
x = 9 / 0.12
x ≈ 75

So, approximately 75 kg of the 24% nitrogen fertilizer should be mixed with 25 kg of the 12% nitrogen fertilizer to obtain 100 kg of fertilizer that is 21% nitrogen.