Question

I need help on question 7 and 8

Answer 1

7.
remember
(ab)/(cd)=(a/c)(b/d)
we can split them up
and
(x^m)/(x^n)=x^(m-n)

$\frac{27k^5m^8}{4k^3*9m^2}= \frac{27k^5m^8}{36k^3m^2}=( \frac{27}{36} )( \frac{k^5}{k^3} )( \frac{m^8}{m^2} )$=$( \frac{3}{4})(k^{5-3})(m^{8-2} )=( \frac{3}{4})(k^2)(m^6 )= \frac{3k^2m^6}{4}$



9.

2 to 3
4(x-1)=15
to
4x-1=15
the distribuutive prperty
a(b-c)=ab-ac
what he did was
a(b-c)=ab-c

he did not distribute the 4 to the -1


4(x-1)+3=18
minus 3
4(x-1)=15
distribute  4 to x and -1
4x-4=15
add 4 to both sides
4x=19
divide by 4
x=19/4