*The problem is asking for both the numbers, so we can make “\(n\)” the smaller number, and “\(18-n\)” the larger.*

Also, remember that if the problem calls for a pure solution or concentrate, use 100%.

Solution: Converting repeating decimal to fraction problems can be easily solved with a little trick; we have to set it up as a subtraction, so the repeating part of the decimal is gone.

Probably the most common is to set up a proportion like we did here earlier. There’s another common way to handle these types of problems, but this way can be a little trickier since the variable in the equation is not what the problem is asking for; we will make the variable a “multiplier” for the ratio.

The advantage to this way is we don’t have to use fractions. We can find out how much of ingredients a and b are in solution X by using a ratio multiplier again (one ounce of solution X contains ingredients a and b in a ratio of Let \(x=\) what you need to make on the final.

Here’s an example of a Quadratic Inequality word problem.

We’ll also use inequalities a lot in the Introduction to Linear Programming section.For example, “ \(\displaystyle \begin\left( \right)n-3=2\left( \right)-33\\,\,\,\,\,-7n-3=-2n-33\\,\,\,\,\,\,\underline\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3\,=\,\,\,5n-33\\,\,\,\,\,\,\,\,\,\,\,\,\,\underline\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,30\,\,=\,\,\,5n\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac\,\,=\,\,\frac\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n=6\end\) \(\displaystyle \beginx=$20 \left( \right)\x=$20 \left( \right)\x=$20 $3=$23\x=$23\,\end\) or \(\displaystyle \beginx=$20\times \left( \right)\x=$20\times \left( \right)\x=$20\times \left( \right)\x=$23\,\end\) Solution: This problem seems easy, but you have to think about what the problem is asking.When we are asked to relate something to something else, typically we use the last thing (the “to the” part) as the \(y\), or the dependent variable.The problems here only involve one variable; later we’ll work on some that involve more than one.Doing word problems is almost like learning a new language like Spanish or French; you can basically translate word-for-word from English to Math, and here are some translations: Note that most of these word problems can also be solved with Algebraic Linear Systems, here in the Systems of Linear Equations section.We can just put a negative sign in front of the variable.If you’re not sure if you should multiply, add, or subtract, try “real numbers” to see what you should do.So the equation relating the number of color photos \(p\) to the number of minutes \(m\) is \(\displaystyle m=\fracp\). Let’s see how we can set this up in an equation, though, so we can do the algebra!There are actually a couple of different ways to do this type of problem.The problem is asking for a number, so let’s make that \(n\).Now let’s try to translate word-for-word, and remember that the “opposite” of a number just means to make it negative if it’s positive or positive if it’s negative.

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