A chemist has three different acid solutions.
Let x denotes the first acid, y denotes the second acid, and z denotes the third acid.
The first solution contains 20% acid the second contains 30% acid and the third contains 50% acid
She created 60 liters of a 40% acid mixture using all three solutions.
Mathematically,
[tex]\begin{gathered} 0.20x+0.30y+0.50z=0.40\cdot60 \\ 0.20x+0.30y+0.50z=24\qquad eq.1 \end{gathered}[/tex]
The number of liters of 50% solution used is three times the number of liters of 30% solution used.
Mathematically,
[tex]z=3\cdot y\qquad eq.2[/tex]
Also, we know that the sum of all solutions must be equal to 60 liters,
Mathematically,
[tex]x+y+z=60\qquad eq.3[/tex]
Now let us substitute the eq.2 into the eq.1
[tex]\begin{gathered} 0.20x+0.30y+0.50z=24 \\ 0.20x+0.30y+0.50(3\cdot y)=24 \\ 0.20x+0.30y+1.5y=24 \\ 0.20x+1.8y=24\qquad eq.4 \end{gathered}[/tex]
Now let us substitute the eq.2 into the eq.3
[tex]\begin{gathered} x+y+z=60 \\ x+y+3y=60 \\ x+4y=60\qquad eq.5 \end{gathered}[/tex]
As you can see, we now have 2 equations with 2 unknowns
Let us solve them using the substitution method.
Re-arrange eq.5 for x and substitute it into eq.4
[tex]\begin{gathered} x+4y=60 \\ x=60-4y \end{gathered}[/tex]
So eq.4 becomes
[tex]\begin{gathered} 0.20x+1.8y=24 \\ 0.20(60-4y)+1.8y=24 \\ 12-0.80y+1.8y=24 \\ 12+y=24 \\ y=24-12 \\ y=12 \end{gathered}[/tex]
So y = 12 liters.
Now let us substitute y = 12 into eq.5
[tex]\begin{gathered} x+4y=60 \\ x+4(12)=60 \\ x+48=60 \\ x=60-48 \\ x=12 \end{gathered}[/tex]
So x = 12 liters
Finally, substitute y = 12 into eq. 2
[tex]\begin{gathered} z=3\cdot y \\ z=3\cdot12 \\ z=36 \end{gathered}[/tex]
So z = 36 liters
Therefore, the chemist used 12 liters of 20% solution, 12 liters of 30% solution, 36 liters of 50% solution.