A 30% solution of fertilizer is to be mixed with a 60% solution of fertilizer in order to get 150 gallons of 50% solution. How many gallons of the 30% and the 60% solutions should be mixed. I really need help.

x = gal of 30% solution

.30x = value of 30% solution
150 - x = gal of 60% solution
.60(150 - x)
.50(150) = mixture

.30x + .60(150 - x) = .50(150)

Solve for x, gal of 30% solution
150 - x = gal of 60% solution

To solve this problem, we can follow a simple approach.

Let's assume that x gallons of the 30% solution is mixed with y gallons of the 60% solution.

We can now set up the two equations based on the information provided:

Equation 1: The total volume of the mixture is 150 gallons: x + y = 150

Equation 2: The total fertilizer concentration in the mixture is 50%: (0.30x + 0.60y)/(x + y) = 0.50

Now, we can solve this system of equations to find the values of x and y.

First, let's simplify equation 2 by multiplying both sides of the equation by (x + y):

0.30x + 0.60y = 0.50(x + y)

0.30x + 0.60y = 0.50x + 0.50y

Now, rearrange the terms:

0.60y - 0.50y = 0.50x - 0.30x

0.10y = 0.20x

Divide both sides by 0.20:

0.10y/0.20 = 0.20x/0.20

0.5y = x

Now substitute this value of x in equation 1:

0.5y + y = 150

1.5y = 150

Divide both sides by 1.5:

y = 150/1.5

y = 100

Now, substitute this value of y back into equation 1 to find x:

x + 100 = 150

x = 150 - 100

x = 50

Therefore, you should mix 50 gallons of the 30% solution and 100 gallons of the 60% solution to get 150 gallons of a 50% solution.

To solve this problem, let's assume we need x gallons of the 30% solution and y gallons of the 60% solution.

Given:
1. The total volume of solution is 150 gallons.
2. The desired concentration of the solution is 50%.
3. The concentration of the 30% solution is 30%.
4. The concentration of the 60% solution is 60%.

Now, we can set up the following equation based on the volume and concentration of the two solutions:

0.30x + 0.60y = 0.50(150)

Let's simplify the equation:

0.30x + 0.60y = 75

Since the total volume is 150 gallons, we have another equation:

x + y = 150

Now, we have a system of two equations:

0.30x + 0.60y = 75
x + y = 150

We can solve this system of equations using the method of substitution or elimination:

Using the substitution method:
1. Solve one of the equations for one variable (y or x) in terms of the other variable.
Let's solve the second equation for x:
x = 150 - y

2. Substitute the expression for the variable obtained in step 1 into the other equation.
0.30(150 - y) + 0.60y = 75

3. Simplify and solve for y:
45 - 0.30y + 0.60y = 75
0.30y = 75 - 45
0.30y = 30
y = 30 / 0.30
y = 100

4. Substitute the value of y back into one of the original equations to solve for x:
x + 100 = 150
x = 150 - 100
x = 50

Therefore, you would need to mix 50 gallons of the 30% solution with 100 gallons of the 60% solution to obtain 150 gallons of a 50% solution.