Translate into 2 linear equations?

[justan4cat]

In a chemistry class, 5 liters of a 4% saline solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed?

I'm thinking that the 4% saline solution=x plus a 10% solution=y to get (=) a 6% solution? But that's only one equation. I don't see two.

If I can get this set up, I can use the elimination or substitution method to solve.

 

[galactus]

We mix 5 liters of 4% and x liters of 10%. This gives 5+x liters.

\(\displaystyle .04(5)+.10x=.06(5+x)\)

Two equations not required.

 

[justan4cat]

The directions for the problem say, "Translate the problem into a pair of linear equations in two variables. Solve the equations using either the elimination or substitution method. State your answer for the specified variable."

That's why I made the subject of the thread, "Translate into 2 linear equations".

 

[justan4cat]

Do you agree with this?

x=liters of 10% solution, Y=total liters of 6% solution

5+x=y
.04(5)+.1(x)=.06(y) Then multiply by 10 to clear fractions
20+10x=6y

Substitution:
20+10(y-5)=6y
20+10y-50=6y
10y-30=6y
4y=30
y=7 1/2

Then:
x=7 1/2-5
x=2 1/2

Answer: 2 1/2 liters of 10% solution are needed to obtain a 6% solution when mixed with 5 liters of a 4% solution.

 

[Deleted member 4993]

Do you agree with this?

x=liters of 10% solution, Y=total liters of 6% solution

5+x=y
.04(5)+.1(x)=.06(y) Then multiply by 10 to clear fractions<<< You are multiplying by 100 - otherwise excellent work
20+10x=6y

Substitution:
20+10(y-5)=6y
20+10y-50=6y
10y-30=6y
4y=30
y=7 1/2

Then:
x=7 1/2-5
x=2 1/2

Answer: 2 1/2 liters of 10% solution are needed to obtain a 6% solution when mixed with 5 liters of a 4% solution.