Solving Systems of Equations: 30%, 55%, 50% Acid Mixture

[opticaltempest]

Hello, I am stuck on the following problem.

How many milliliters of a 30% acid solution and a 55% acid solution must be mixed to obtain 100 milliliters of a 50% acid solution?

My solution:

Let X = The unknown milliliters of 30% acid.
Let Y = The unknown milliliters of 55% acid.

(1) X + Y = 100
(2) 0.3X + 0.55Y = 0.5

Multiplying (2) by 100 in order to work with integers,

(2) 30X + 55Y = 50

Multiplying (1) by -30 in order to eliminate X

(1) -30X - 30Y = -3000

So our new equations are,

(1) -30X - 30Y = -3000
(2) 30X + 55Y = 50

Adding the two equations in order to eliminate X,

25Y = -2950 => Y = -118

Back-substituting Y into (1) in order to solve for X,

-30X - 30(-118) = -3000

-30X + 3540 = -3000

-30X = -6540

X = 218

The answer in the book says:

20 milliliters of 30% acid solution and 80 milliliters of 55% acid solution.Any hints as to where I am going wrong?

 

[vaishakh]

(2) 0.3X + 0.55Y = 0.5

The above part is wrong. the RHS is wrong. think?

 

[Quadratic]

How many milliliters of a 30% acid solution and a 55% acid solution must be mixed to obtain 100 milliliters of a 50% acid solution?

For starters, let's get the right first 2 equations. I hope you understand the following:

50 = .3x + .55y
100 = x + y

y = 100 - x
50 = .3x + .55 (100 - x)
50 = .3x + 55 - .55x
-5 = -.25x
(-5/-.25) = x
x = 20
100 - 20 = y
y = 80

So it's 20 mills of 30%, and 80 mills of 55%