This topic focus is about solving word problems. Below are many examples from the Linear Things unit involving word problems. In parenthesis is a key word from a topic alread discussed should you need to go back for some help.
Amber is signing up for cell phone service and needs to decide between two different plans. Plan A costs $54.99 a month with a free-phone included, while Plan B costs $49.99 a month, but requires her to buy a phone for $129. Under either plan, Amber does not expect to go over the included number of monthly minutes. After how many months would Plan B be a better deal?
Let \(x\) represent the number of months and \(C\) the monthly cost.
| Plan A: \(C=54.99x\) | Plan B: \(C=49.99x+129\) | Plans B better deal \(\implies 54.99x \geq 49.99x+129\) |
| \(\implies 5x \geq 129 \implies x \geq \frac{129}{5} \approx 25.8\) | ||
| After 26 months, Plan B would be a better deal. | ||
In a supply and demand system for frozen yogurt, when the price is $2 the number of half-gallons in demand is 400. However, at $2, the supply is only 200. The demand drops to 100 half-gallons at $8, while the supply is at 500 half-gallons. Create a system of linear equations that help determine the price when the supply equals the demand. How many half-gallons will that price allow?
Let \(x\) be the price and \(y\) the number of half-gallons and we can build a linear equation for the supply and for the demand:
| Supply: \((2,200)\) and \((8,500)\) | Demand: \((2,400)\) and \((8,100)\) |
| \(m=\frac{200-500}{2-8}\implies m=\frac{-300}{-6}\implies m=50\) | \(m=\frac{400-100}{2-8}\implies m=\frac{300}{-6}\implies m=-50\) |
| Point-slope: \(y-200=50\left(x-2\right)\) | Point-slope: \(y-400=-50\left(x-2\right)\) |
| System (in \(y=\) form): | \(\begin{array}{l} y=50\left(x-2\right)+200 \\ y=-50\left(x-2\right)+400\end{array}\) |
| Substitution: | \(50\left(x-2\right)+200=-50\left(x-2\right)+400\) |
| \(50x-100+200=-50x+100+400\) | |
| \(100x=400\implies x=4\) |
Supply equals demand at the price of $4. Now plug back into either equation in the system to find the number of half-gallons, \(y\).
$$y=50(4-2)+200 \implies y=50(2)+200 \implies y=300$$Hence, the number of half-gallons will be 300.
How many ounces each of \(5\%\) hydrochloric acid and \(20\%\) hydrochloric acid must be combined to get \(10\) oz of a solution that is \(12.5\%\) hydrochloric acid?
| Let \(x\) represent oz of \(5\%\) solution | Let \(y\) represent oz of \(20\%\) solution |
| TOTAL number of ounces equation: | \(x+y=10\) |
| Solution equation: | \(0.05x+0.20y=0.125(10)\) |
| Elimination: (multiply ounces equation by \(-0.20\)) | \(-0.20x-0.20y=-2\) |
| Add together to eliminate \(y\)s: | \(\begin{array}{c} -0.20x-0.20y=-2 \\ \underline{+0.05x+0.20y=1.25} \end{array}\) |
| \(-0.15x=-0.75\implies x=5\) | |
| Substitue back in \(x=5\): | \(5+y=10\implies y=5\) |
So \(5\) ounces each of the \(5\%\) and the \(20\%\) solutions are needed.
Soccer fields can have dimensions that vary. The width can be between \(50\) and \(100\) yds, and the length between \(100\) and \(130\) yds. Suppose one field has a perimeter of \(320\) yds, and the length is \(40\) yds more than the width. Find the dimensions of this soccer field.
| Write two equations: | Perimeter is \(320\): \(\implies 2W+2L=320\) |
Length is \(40\) yds more than width: \(\implies L=W+40\) |
| Solve using any method: | Substitution works well here: | \(2W+2(\color{blue}{W+40})=320\) \(\implies 2W+2W+80=320\) \(\implies 4W+80\color{red}{-80}=320\color{red}{-80}\) \(\implies \large{\frac{4W}{\color{red}{4}}}\normalsize{=}\large{\frac{240}{\color{red}{4}}}\) \(\implies W=60\) |
| Plug back in to solve for length: | \(\implies L=\color{blue}{60}+40\) \(\implies L=100\) |
So the dimensions of the soccer field must be \(40\) yds wide by \(60\) yds long.