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1. Roger is a wildlife researcher who has been hired to study and track the population of bears in a national park. In Year 1 of the study, Roger determines that the park is home to x bears. In Year 2, the population has increased to x+10 bears, and in Year 3, the population has increased to 2(x+10) bears.

A. Use the distributive property to rewrite the bear population in Year 3 without needing to use parenthesis

B. The total number of bears in the park during Year 2 is 42. Write and solve a linear equation to determine how many bears were in the park during Year 1. Explain the answer

C. The total number of bears in the park during Year 3 is 50. Write and solve a linear equation to determine how many bears were in the park during Year 1. Note that this answer will not be the same as the answer you got in part b.

D. During Year 4, bad weather keeps Roger from finding the exact number of bears in the park. However he knows that y+5>38, where y is the number of bears in the park during Year 4. Solve the inequality, and give three possible values for y.

1 answer

Let’s solve each part step-by-step.

A. We start with the bear population in Year 3, which is given as \( 2(x + 10) \). Using the distributive property, we can rewrite this without parentheses:

\[ 2(x + 10) = 2x + 20 \]

So, the bear population in Year 3 can be expressed as \( 2x + 20 \).

B. We know that the total number of bears in the park during Year 2 is 42. The number of bears in Year 2 is given by \( x + 10 \). We can set up the following equation:

\[ x + 10 = 42 \]

Now, we solve for \( x \):

\[ x = 42 - 10 = 32 \]

This means that during Year 1, Roger found that there were \( 32 \) bears in the park.

Explanation: Since Year 2 has \( 10 \) more bears than Year 1, we can verify that if Year 1 had \( 32 \) bears then Year 2 would indeed have \( 32 + 10 = 42 \) bears.

C. The total number of bears in the park during Year 3 is 50. We express the number of bears in Year 3 as \( 2x + 20 \). We can set up the following equation:

\[ 2x + 20 = 50 \]

Now, we solve for \( x \):

  1. Subtract \( 20 \) from both sides:

\[ 2x = 50 - 20 \ 2x = 30 \]

  1. Divide both sides by \( 2 \):

\[ x = \frac{30}{2} = 15 \]

Thus, during Year 1, the number of bears was \( 15 \).

Explanation: This difference arises because the population increased differently in Years 2 and 3. Here, if Year 1 had \( 15 \) bears, then Year 2 would have \( 15 + 10 = 25 \) bears, which satisfies the Year 3 calculation of \( 2(25) = 50 \) bears.

D. We need to solve the inequality \( y + 5 > 38 \) to determine possible values for \( y \).

  1. Subtract \( 5 \) from both sides:

\[ y > 38 - 5 \ y > 33 \]

Now, we can give three possible values for \( y \) that satisfy this inequality. For example:

  • \( y = 34 \)
  • \( y = 35 \)
  • \( y = 40 \)

These values are all greater than \( 33 \).