Let $\displaystyle y$ = money earned and $\displaystyle x$ = number of cars sold

So $\displaystyle y = 500x + 1000$

$\displaystyle 7500 = 500x + 1000$ and solving shows that he needs to sell 13 cars to make $7,500.

Let pure water = 0 % and pure acid = 100%

The general equation to use is:

$\displaystyle V_{1}P_{1} + V_{2}P_{2} = V_{f}P_{f}$  where $\displaystyle V$is the volume and $\displaystyle P$is the percent solution.

So the equation to solve becomes $\displaystyle x(0) + 2(.80) = (x + 2)(.20)$ and $\displaystyle x = 6$ gallons of pure water needs to be added to get a 20% acid solution.

The break-even point is where the costs equal revenue.

Let $\displaystyle w$ = # of widgets sold.

Costs:  $\displaystyle C(w) = 45w + 750$

Revenue:  $\displaystyle R(w) = 75w$

So the equation to solve becomes $\displaystyle 45w + 750 = 75w$

So the break-even point occurs when they sell 25 widgets.

Profits = Revenues - Costs

Revenue:  $\displaystyle R(w) = 75w$

Costs:  $\displaystyle C(w) = 45w + 750$

Profit: $\displaystyle P(w) = 75w - (45w + 750) = 30w -750$

So the equation to solve becomes $\displaystyle 3,000 = 30w - 750$

So a $3,000 profit occurs when they sell 125 widgets

The break-even point is where the costs equal the revenues.

Let $\displaystyle x$ = # of frames sold

Costs:  $\displaystyle C(x) = 10x +350$

Revenues:  $\displaystyle R(x) = 35x$

Thus, $\displaystyle 10x + 350 = 35x$

So 14 picture frames must be sold each month to break-even.

Let $\displaystyle x$ = # of frames sold

$\displaystyle Profits = Revenues-Costs$

Revenues:  $\displaystyle R(x) = 35x$

Costs:  $\displaystyle C(x) = 10x + 350$

Profits = $\displaystyle P(x) = R(x) - C(x) = 35x - (10x + 350) = 25x - 350$

So the equation to solve becomes $\displaystyle 500 = 25x - 350$

So 34 picture frames must be sold to make a $500 profit.

Let pure water be 0% and pure solution be 100%.

So the general equation to solve is:

$\displaystyle V_{1}P_{1} + V_{2}P_{2} = V_{f}P_{f}$ where $\displaystyle V$ is the volume and the $\displaystyle P$ is percent solution.

So the equation to solve becomes $\displaystyle x(0) + 2(0.90) = (x + 2)(0.30)$

Solving shows that we need to add 4 gallons of pure water to 2 gallons of 90% pure cleaning solution to get a 30% pure solution.

Let $\displaystyle x$ = number of dimes, $\displaystyle x + 1$ = number of nickels, and 

$\displaystyle (x +1) - 2 = x - 1$ = number of quarters.

The general equation to use is:

$\displaystyle V_{1}N_{1} + V_{2}N_{2} + V_{3}N_{3} = V_{f}$ where $\displaystyle V$ is the money value and $\displaystyle N$ is the number of coins

So the equation to solve becomes

$\displaystyle 0.10x + 0.05(x + 1) + 0.25(x - 1) = 1.40$

Thus, solving the equation shows that she had five nickels, four dimes, and three quarters giving a total of 12 coins.

Pure water is 0% and pure solution is 100%

$\displaystyle V_{1}P_{1} + V_{2}P_{2} = V_{f}P_{f}$ where $\displaystyle V$ is the volume and $\displaystyle P$ is the percent.

So the equation to solve becomes $\displaystyle x(0)+1(0.80)=(1+x)(0.25)$

So we need to add $\displaystyle 2.2\ L$ pure water to $\displaystyle 1\ L$ of 80% cleaning solution to yield 25% cleaning solution.

We are looking for the value of t  that gives $675 as the result when plugged in V (t ). While there are many ways to do this, one of the fastest is to plug in the answer choices as values of t .

When we plug t  = 1 into V (t ), we get V (1) = 1200(0.75)¹ = 1000(0.75) = $900, which is incorrect.

When we plug t  = 2 into V (t ), we get V (2) = 1200(0.75)² = $675, so this is our solution.

The value of the tractor will be $675 after 2 years.

Finally, we can see that if t  = 3, 4, or 5, the resulting values of the V (t ) are all incorrect.