150k views
5 votes
4 Algebra problems, can I please get some help? (With steps to solve each one!)

4 Algebra problems, can I please get some help? (With steps to solve each one!)-example-1
User Broschb
by
8.2k points

1 Answer

2 votes
a) a³b² c a
------- x ----- ÷ -----
c²d² ab c²d³

First perform multiplication:

(a³b²c / abc²d²) ÷ a / c²d³

In dividing fractions. Get the reciprocal of the 2nd fraction and multiply it to the 1st fraction.

a/c²d³ is the 2nd fraction. Its reciprocal is c²d³/a

So,
a³b²c c²d³ a³b²c³d³
------------ x --------- = -------------- = abcd *cancel out like terms
abc²d² a a²bc²d²

b) (6x / 4x -16) ÷ (4x / x² -16)
(6x / 4x-16) × (x²-16 / 4x)
6x(x² - 16) / 4x(4x-16)
6x³ - 96 / 16x² - 64x

c) 3x² - 6x x + 3x² *use distributive property of multiplication
--------------- × -------------
3x + 1 x² -4x + 4

3x² (x+3x²) - 6x (x + 3x²) 3x³ + 9x⁴ - 6x² - 18x³
3x (x² - 4x + 4) + 1(x² -4x +4) ⇒ 3x³ -12x² + 12x + x² -4x + 4

9x⁴ + 3x³ - 18x³ - 6x² 9x⁴ - 15x³ - 6x²
3x³ - 12x² + x² + 12x - 4x + 4 ⇒ 3x³ - 11x² + 8x + 4

d) 2x² - 10x + 12 2 + x
---------------------- × -----------
x² - 4 3 - x

2(2x² - 10x + 12) + x (2x² - 10x + 12)4x² - 20x + 24 + 2x³ -10x² + 12x
3(x² - 4) - x(x² -4) 3x² - 12 - x³ + 4x

2x³ + 4x² - 10x² - 20x + 12x + 242x³ - 6x² - 8x + 24
-x³ + 3x² + 4x - 12 -x³ + 3x² + 4x - 12
User Zizoo
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories