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Can someone help me with this problem. I came out with the answer of 7.8 ounces needed. I used x =15% solution and y = 35% solution. I just don't know for sure if I did it right. The problem is:

A 15% acid solution is to be mixed with a 35% acid solution to produce 12 ounces of a 22% acid solution. How much of the 15% acid solution is needed? Let x = ____ Let y = ____ Thank you.

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    $\begingroup$ The $7.8$ is right. So it is very likely that you did it a right way. If $x$ is the amount of $15\%$ solution, and $y$ is the amount of the other kind, we get equations $x+y=12$ and $0.15x+0.35y=(12)(0.22)$. $\endgroup$ – André Nicolas Feb 19 '15 at 5:02
  • $\begingroup$ The amount of acid present before mixing is the same as the amount of acid present after mixing. Your answer, and all of the other answers offered here, follow the same strategy. $\endgroup$ – John Joy Feb 19 '15 at 21:04
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You were right' we need 7.8 ounces of 15% solution and 4.2 ounces of 35% solution to come up with a 12 ounces of 22% acid solution.

The main equation will be:

$.15x+.35(12-x)=12(.22)$ solving for $x$ we get 7.8.

Congrats.

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  • $\begingroup$ Just wondering why if my answer had been down voted :-(. $\endgroup$ – Jr Antalan Feb 19 '15 at 9:59
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    $\begingroup$ While I did not vote down your solution, I would have avoided any reference to the sex of the person who posted the question in my answer. $\endgroup$ – N. F. Taussig Feb 19 '15 at 10:58
  • $\begingroup$ Got it, thanks @N.F.Taussig $\endgroup$ – Jr Antalan Feb 19 '15 at 11:35
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This problem is expressed mathematically as $$ 0.15x+0.35y=12(0.22) $$ $$ x+y=12 $$ This implies that $$ y=12-x $$ So now we have $$ 0.15x+0.35(12-x)=12(0.22) $$ $$ 0.15x+(12)0.35-0.35x=12(0.22) $$ $$ 0.15x+4.2-0.35x=2.64 $$ $$ 0.15x-0.35x=2.64 -4.2$$ $$ -0.2x=-1.56$$ $$ 0.2x=1.56$$ $$ x=\frac{1.56}{0.2}=7.8$$ So yes, your solution is correct.

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Yes, $7.8$ is the correct answer.

Your problem can be modeled by a system of two equations: $$\begin{cases}0.15x+0.35y=12\cdot0.22\\ x+y = 12 \end{cases}$$ To solve it, you can use any of the many techniques available. For example, multiplying the two sides of the second equation by $0.15$: $$\begin{cases}0.15x+0.35y=12\cdot0.22\\ 0.15x+0.15y = 12\cdot0.15 \end{cases} \stackrel{subtracting}{\implies}\begin{cases}y = 5\cdot12\cdot0.07\\ x + y = 12 \end{cases}\implies \begin{cases}y = 4.2\\ x = 7.8 \end{cases}$$

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