*If Erica works let’s say Since there is a one-time cost in addition to a per-person cost, the cost per person will depend on the number of students attending the party: the more students, the lower the cost.For \(x\) students attending, each would have to pay \(\displaystyle \frac\) for the bowling alley rent; try it with real numbers!Now we have 6 test grades that will count towards our semester grade: 4 regular tests and 2 test grades that will be what you get on the final (since it counts twice, we need to add it HINT: For any problem with weighted averages, you can multiply each value by the weight in the numerator, and then divide by the sum of all the weights that you’ve used.*

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.

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.

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

We’ll also use inequalities a lot in the Introduction to Linear Programming section.

To do this, let \(x=\) the repeating fraction, and then we’ll figure out ways to multiply \(x\) by ) just to the right of the decimal point; we get \(10x=4.\underline2525…\).

Now we have to line up and subtract the two equations on the left and solve for \(x\); we get \(\displaystyle x=\frac\). Let’s see if it works: Put \(\displaystyle \frac\) in your graphing calculator, and then hit Enter; you should something like \)),“is no more than” (\(\le \)), “is at least” (\(\ge \)), and “is at most” (\(\le \)).

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.

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