Question

if ab=3x+2, bc=8x-10, ac=69, and x>?

Answer 1

what you want to do?

Answer 2

Show a ss please

Answer 3

ab=3x+2
abc=c(3x+2)...(1)
bc=8x-10
abc=a(8x-10) ...(2)
from (1) and (2)
a(8x-10)=c(3x+2)
ac=69
\[a=\frac{ 69 }{ c }\]
\[\frac{ 69 }{ c }(8x-10)=c(3x+2)\]
\[\frac{ 69(8x-10) }{ 3x+2 }=c^2\]
\[c^2\ge 0~(always)\]
\[Hence~\frac{ 69(8x-10) }{ 3x+2 }\ge 0\]
or
\[\frac{ 8x-10 }{ 3x+2 }\ge 0\]
so numerator and denominator are of same sign.
\[let~8x-10 \ge 0\]
or
\[x \ge \frac{10}{8}\]
\[x\ge \frac{5}{4}\]
and \[3x+2 \ge 0\]
\[x\ge \frac{-2}{3}\]
combining \[x \ge \frac{5}{4}\]
similarly you can solve for 8x-10<0
and 3x+2<0

Answer 4

8x-10<0
\[x <\frac{ 10 }{ 8 }\]
or
\[x<\frac{ 5 }{4 }\]
and
3x+2<0
\[x<\frac{ -2 }{3 }\]
combining
\[x<-\frac{ 2 }{3 }\]
combinig
\[x<-\frac{ 2 }{ 3 } ~or~x \ge \frac{ 5 }{ 4 }\]