Please help with these algebra problems?

Question

1. |t-7| -5 = 4 

2. |2a+4| = |3a-1| 

3. |7-2y| > 5 

4. |t-7| + 3 ≥ 4 

Please explain how to solve! Thanks!

Answer 1

1).

The equation is |t - 7| - 5 = 4.

First rewrite the above equation in the form |ax + b| = c where c  ≥ 0 and it is equivalent to the statement  ax + b = c or ax + b = - c.

|t - 7| - 5 + 5 = 4 + 5

|t - 7| = 9

t - 7 = ± 9

t - 7 = 9  and   t - 7 = - 9

t = 9 + 7  and  t = - 9 + 7

t = 16 and t = - 2.

The solutions of the equation are t = 16 and t = - 2.

2).

The equation is |2a + 4| = |3a - 1|.

If | a | = | b |, then a = b.

2a + 4 = 3a - 1

2a - 3a = - 1 - 4

- a = - 5

a = 5.

The solution of the equation is a = 5.

3).

The absolute value inequality is |7 - 2y| > 5.

The absolute value inequality |7 - 2y| > 5 is equivalent to 7 - 2y > 5 or 7 - 2y  < - 5.

Solve the inequality 1 : 7 - 2y > 5

- 2y > 5 - 7

- 2y > - 2

y < 1.

Solve the inequality 2 : 7 - 2y  < - 5

 - 2y  < - 5 - 7

 - 2y  < - 12

y > 6.

The solution set of the inequality is { y | y : y > 6 or y < 1 }.

4).

The absolute value inequality is |t - 7| + 3 ≥ 4.

|t - 7| + 3 ≥ 4 = |t - 7| ≥ 4 - 3 = |t - 7| ≥ 1.

The absolute value inequality |t - 7|  ≥ 1 is equivalent to (t - 7 ) ≥ 1 or (t - 7)  ≤  - 1.

Solve the inequality 1 : (t - 7 ) ≥ 1

t - 7  ≥ 1

t  ≥ 1 + 7

t  ≥ 8.

Solve the inequality 2 :(t - 7) ≤  - 1

t ≤  - 1 + 7

t  ≤ 6.

The solution set of the inequality is { t | t :t  ≥ 8 or t  ≤ 6 }.