I’m having a problem trying to solve this Monomial math question?

2.

-14ab^2c/21a^2bc^2

= -2b/3ac

5,

c^3d^2/c^8d^4

= 1/c^5d^2

x^a / x^b = x^(a-b)

a/a^2 = a^(1-2) = a^-1 = 1/a

b^2 / b = b^(2-1) = b

c/c^2 ... like the first = 1/c

14 = 2*7

21 =3*7

the answer = (-2/3) b/(a*c)

the other's answer = 1/(c^5 * d^2)

When dividing monomials, you can rewrite the big division as a multiplication of tiny divisions, each one grouping only similar objects.

Here, you have 4 types of objects:

numbers

"a"

"b"

"c"

-14ab^2c / 21a^2bc^2

can be rewritten as

(-14/21)(a/a^2)(b^2/b)(c/c^2)

Each one, then, can be solved by cancelling appropriately.

If you want to do with with "baby steps" (at first), factor everything:

(-14 / 21) = -(7*2 / 7*3)

the "7" cancels out, leaving -2/3

(a / a^2) = (a / a*a)

the a on top is the same as a*1

(multiplying by 1 never changes a value)

(a*1 / a*a)

we can cancel one "a" above with one "a" below, leaving us with

(1/a)

(b^2 / b) = (b*b / b*1) = (b/1) = b

(since dividing by 1 changes nothing, we don't bother to write it).

(c / c^2) = (c*1 / c*c) = (1/c)

Now, we rebuild the answer by multiplying the simplified factors:

(-2/3)(1/a)(b/1)(1/c)

top with top, bottom with bommon

-2*1*b*1 / 3*a*1*c

ignore the "1"s

-2b / 3ac

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faster trick

(knows as the law of exponents)

When you multiply powers of the same base, add the powers

a^3 * a^4 = (aaa)(aaaa) = aaaaaaa = a^7

a^3 * a^4 = a^(3+4) = a^7

When you divide power, subtract:

a^5 / a^2 = aaaaa / aa = (aaa)(aa) / (aa) = aaa = a^3

a^5 / a^2 = a^(5-2) = a^3

Applying this, you will find that a^0 = 1

Anything divided by itself is equal to 1 (except 0/0, but we don't care about that today)

a^3 / a^3 = 1 (something divided by itself

a^3 / a^3 = a^(3-3) = a^0

therefore, a^0 = 1

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A fraction is the same as a negative power.

1/a^2 = a^0 / a^2 = a^(0-2) = a^(-2)

When you get a negative power, you can leave it there (with the negative sign in the power) or you can rewrite it as a fraction. Remember that "a" by itself is the same as a^1

a^(-1) is the same as (1/a)

Therefore, in your problem:

a/a^2 = a^1 / a^2 = a^(1-2) = a^(-1) = (1/a)

Sounds complicated, but you get the hang of it very rapidly.

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Problem number 5:

c^3 / c^8 = c^-5 = 1 / c^5

d^2 / d^4 = d^-2 = 1 / d^2

= 1 / (c^5d^2)