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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
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1 Answer

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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
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