i need help working these correctly!

1. Are the inequalities 3x-4<10-4x and 2(x-5) >3(2x-6) equivalent? Why or why not? Show your work and explain your answer.

3. The yearly cost T, in thousands of dollars, of tuition and required fees at a private college (includes two- and four-year schools and does not include room and board) can be approximated by

T = 3/5n+5


where n is the number of years since 1985. That is, corresponds to 1985, corresponds to 1990, and so on. Show your work and explain each answer.
a. Find the cost of tuition in 1985, 1996, 2000, and 2004.estimate the cost of tuition in 2005.Predict the year in which the cost of tuition will be $23,000. Explain your answer.

1. 3X-4<10-4X. 2(X-5)>3(2X-6)

3X-4<10-4X, 7X<14, X<2.

2(X-5)>3(2X-6), 2X-10>6X-18, -4X>-8,
X<2.

Since the solution to both inequalities is the same(X<2),they
are equivalent.

1. To determine if the inequalities are equivalent, we need to solve them and compare the solutions.

For the first inequality:
3x - 4 < 10 - 4x

We start by getting all the x terms on one side:
3x + 4x < 10 + 4

Combining like terms:
7x < 14

Finally, we divide both sides by 7 to solve for x:
x < 2

Now, let's solve the second inequality:

2(x - 5) > 3(2x - 6)

Start by distributing the 2 and 3:
2x - 10 > 6x - 18

Get all the x terms on one side:
2x - 6x > -18 + 10

Combine like terms:
-4x > -8

Divide both sides by -4 (note that we flip the inequality sign because we are dividing by a negative number):
x < 2

Comparing the solutions, we see that both inequalities have the same solution: x < 2. Therefore, the two inequalities are equivalent.

2. To solve for the yearly cost of tuition, we substitute the given values into the formula T = (3/5)n + 5.

a. Cost of tuition in 1985 (n = 0):
T = (3/5)(0) + 5
T = 0 + 5
T = 5,000

Cost of tuition in 1996 (n = 11):
T = (3/5)(11) + 5
T = 33/5 + 5
T = 6.6 + 5
T ≈ 11.6
The cost of tuition in 1996 is approximately $11,600.

Cost of tuition in 2000 (n = 15):
T = (3/5)(15) + 5
T = 45/5 + 5
T = 9 + 5
T = 14
The cost of tuition in 2000 is $14,000.

Cost of tuition in 2004 (n = 19):
T = (3/5)(19) + 5
T = 57/5 + 5
T = 11.4 + 5
T ≈ 16.4
The cost of tuition in 2004 is approximately $16,400.

Next, we estimate the cost of tuition in 2005 (n = 20):
T = (3/5)(20) + 5
T = 60/5 + 5
T = 12 + 5
T = 17
The estimated cost of tuition in 2005 is $17,000.

Finally, to predict the year in which the cost of tuition will be $23,000, we set T = 23 and solve for n:
23 = (3/5)n + 5
18 = (3/5)n
90 = 3n
n = 30

Therefore, the cost of tuition will be $23,000 in the year 2015 (1985 + 30).