a college lab paced two orders with a chemical supplier. the first order was for 13 bottles of calcium hydroxide solution and 4 bottles of sulfuric acid and totalled $ 487. the second order was for 6 bottles of calcium hydroxide solution and 2 bottles of sulfuric acid and totalled $232.the bills do not list per item price.what were the costs of one bottle of calcium hydroxide solution ad one bottle of sulfuric acid?

To determine the cost of one bottle of calcium hydroxide solution (CHS) and one bottle of sulfuric acid (SA), we can set up a system of equations using the information from the two orders.

Let's assume the cost of one bottle of calcium hydroxide solution (CHS) is represented by x, and the cost of one bottle of sulfuric acid (SA) is represented by y.

From the first order:
13x + 4y = 487

From the second order:
6x + 2y = 232

To solve this system of equations, we can use the substitution method:

1. Solve one of the equations for one variable:
Let's solve the second equation for x:
6x = 232 - 2y
x = (232 - 2y) / 6

2. Substitute the value found into the other equation:
13((232 - 2y) / 6) + 4y = 487

Now, let's solve this equation step by step:

Step 1: Multiply 13 by ((232 - 2y) / 6):
(13/6)(232 - 2y) + 4y = 487

Step 2: Distribute (13/6) to (232 - 2y):
(13/6)(232) - (13/6)(2y) + 4y = 487

Step 3: Simplify and combine like terms:
(3046/6) - (26/6)y + 4y = 487
(509 - 26y + 24y)/6 = 487

Step 4: Combine like terms in the numerator:
(509 - 2y)/6 = 487

Step 5: Multiply both sides of the equation by 6 to eliminate the denominator:
509 - 2y = 487 * 6

Step 6: Simplify and solve for y:
509 - 2y = 2922
-2y = 2922 - 509
-2y = 2413
y = 2413 / -2
y = -1206.5

Now that we have the value of y, we can substitute it back into either of the original equations to find the value of x.

Let's substitute it into the second equation:
6x + 2(-1206.5) = 232
6x - 2413 = 232
6x = 232 + 2413
6x = 2645
x = 2645 / 6
x ≈ 440.83

Therefore, the cost of one bottle of calcium hydroxide solution (CHS) is approximately $440.83, and the cost of one bottle of sulfuric acid (SA) is approximately $1206.50.

To solve this problem, we need to set up a system of equations using the given information. Let's represent the cost of one bottle of calcium hydroxide solution as "x" and the cost of one bottle of sulfuric acid as "y".

From the first order, we know that 13 bottles of calcium hydroxide solution and 4 bottles of sulfuric acid totaled $487. This can be written as:

13x + 4y = 487

From the second order, we know that 6 bottles of calcium hydroxide solution and 2 bottles of sulfuric acid totaled $232. This can be written as:

6x + 2y = 232

Now, we have a system of two equations:

13x + 4y = 487
6x + 2y = 232

We can solve this system of equations either by substitution or elimination method.

Using the elimination method, we can multiply the second equation by 2 to make the coefficients of "y" the same. This gives us:

13x + 4y = 487
12x + 4y = 464

Now, we can subtract the second equation from the first equation:

(13x + 4y) - (12x + 4y) = 487 - 464
13x - 12x = 23

Simplifying this equation, we get:

x = 23

Now, we can substitute this value of x back into one of the original equations to find the value of y. Let's use the second equation:

6(23) + 2y = 232
138 + 2y = 232
2y = 232 - 138
2y = 94
y = 94/2
y = 47

Therefore, the cost of one bottle of calcium hydroxide solution is $23 and the cost of one bottle of sulfuric acid is $47.