a. 2(x + 10) = 2x + 2(10) = 2x + 20
b. Let the number of bears in year 1 be represented by the variable y. We are given that in year 2, the population is 42 bears, so we can set up the equation:
y + 10 = 42
y = 42 - 10
y = 32
Therefore, in year 1 there were 32 bears in the park.
c. Let y represent the number of bears in year 1. We are given that in year 3, the population is 50 bears, so we can set up the equation:
2(y+10) = 50
2y + 20 = 50
2y = 50 - 20
2y = 30
y = 15
Therefore, in year 1 there were 15 bears in the park.
d. Solve the inequality y + 5 > 38:
y > 33
Three possible values for y could be 34, 35, or 36.
2.
a. Set up the equation since the number of wolves in year 2 is equal to the number of wolves in year 3:
7 + z = 3z - 3
7 + 3 = 3z - z
10 = 2z
z = 5
Therefore, in year 1 there were 5 wolves in the park.
b. An expression that could describe the number of wolves in year 4 could be (z + a)(z - b), where a and b are whole numbers greater than 1 and less than 10. One example expression could be (5 + 4)(5 - 2) which simplifies to (9)(3) = 27.