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# How to Solve a Math Problem

There are three steps to solving a math problem.

1. Figure out what the problem is asking.

2. Solve the problem.

3. Check the answer.

### Sample Problem

It's early spring, and Cierra is planting a new lawn. She's decided that the length of the lawn should be 1 ft less than double the width, and the area should be 15 ft2. She also plans to construct a concrete path around the lawn. Fancy. Find the length of the path.

In this problem, we need to find the perimeter of the lawn. Since the dimensions of the lawn are unknown, let l be the length and w be the width. The length (l) is 1 ft less than double the width (2w), or l = 2w – 1.

The area of the lawn is given as 15 ft2, so lw = 15. Let's plug our new equation l = 2w – 1 into that area formula.

(2w – 1)w = 15
2w2w – 15 = 0

Hey, lookie there: it's a polynomial equation. A quadratic, to be specific. We can solve the quadratic equation by factoring:

(2w + 5)(w – 3) = 0

So and 3.

After discarding the negative value, since it would be extra-super-fancy (not to mention bizarre) to have a negative lawn width, we find that w = 3 ft and l = 2(3) – 1 = 5 ft. The length of the path is the perimeter of the lawn:

2(l + w) = 2(5 + 3) = 16 ft

### Sample Problem

Serena and her BFF (or for this week, anyway) Blair want to go to a movie playing at a theater 45 miles away. They're planning to drive separately and meet up at the theater. Whatever, saving on gas is for plebes. Serena starts out first, driving at 40 mph, while Blair starts driving 10 minutes later at 50 mph. When will Blair pass Serena?

(Because she will pass Serena. Hey, nobody ever said they were great drivers.)

Let t be the time in hours when Blair passes Serena. Therefore, the distance traveled by Blair in t hours will be same as the distance traveled by Serena in t hours and 10 minutes (she starts driving 10 minutes, or hr, earlier). Remember the speed formula:

We can rearrange that a bit to get:

distance = speed × time

Using Serena and Blair's driving speed given in the problem, we solve the following linear polynomial equation:

Blair will pass Serena after 40 minutes. Eat her dust, xoxo!

### Sample Problem

Bobby Hill decides to open a factory to produce Shmickerdoodle cookies. He found out that the cost of making q boxes is given by C(q) = 800 + 3q. If he charges \$3.50 per box, how many boxes should the factory produce in order to have no loss?

The revenue obtained by selling q boxes is the price of each box times the number of boxes sold, R(q) = 3.5q. Note that R(q) and C(q) are both polynomials of degree 1. The formula for loss is given by subtracting the cost C(q) from the revenue R(q):

Loss = R(q) – C(q) = 3.5q – (800 + 3q) = 0.5q – 800

The loss is 0 when 0.5q – 800 = 0, which means q = 800 ÷ 0.5 = 1600 boxes.

That's a long way from breaking even, Bobby-o. Hope those Shmickerdoodles are worth it.