Question

EXAM TOMORROW already have work done, just need explanation.
Solve (2x-1)/(x+3)
11 years ago

Answer 1

then (2x-1)/(x+3)-1(x+3)/(x+3)</0

Answer 2

I dont understand where the 1 went?

Answer 3

and i dont get why you subtract x+3

Answer 4

hmmm

Answer 5

\(\large \bf \cfrac{2x-1}{x+3}\le 1?\)

Answer 6

then he put (2x-1-x-3)/(x+3)</0

Answer 7

yes

Answer 8

well.... think about it this way
\(\bf \cfrac{a}{b}\le 1\) if you subtract 1 from both sides.. you'd end up with
\(\bf \cfrac{a}{b}-1\le \cancel{ 1-1 }\implies \cfrac{a}{b}-1\le 0\)

Answer 9

you'd work with an inequality the same way you'd simplify a linear equation
so... there's an assumption you do know how to simplify linear equations

but the procedure is pretty much the same.... with one exception, but that doesn't apply here

Answer 10

the easier way to go about it in this case, is just a cross-multiplication though

Answer 11

why did he subtract x+3 though? and not on the other side?

Answer 12

and if its subtracted on the bottom too, why does x+3 not change on the bottom?

Answer 13

sorry I'm just really confused

Answer 14

well.... that looks odd.... but I see what he did gimme one sec

Answer 15

\(\bf \cfrac{2x-1}{x+3}\le 1\implies \cfrac{2x-1}{x+3}-1\le 0\qquad
\begin{cases}
\frac{\cancel{ same }}{\cancel{ same }}\to 1
\\ \quad \\
\frac{\cancel{ whatever }}{\cancel{ whatever }}\to 1\qquad thus
\\ \quad \\
\frac{\cancel{ (x+3) }}{\cancel{ (x+3) }}\to 1\leftarrow \frac{(x+3) }{(x+3) }
\end{cases}
\\ \quad \\
\cfrac{2x-1}{x+3}-1\le 0\implies \cfrac{2x-1}{x+3}-\cfrac{(x+3) }{(x+3) }\le 0\implies \cfrac{(2x-1)\ -\ (x+3)}{x+3}\le 0
\\ \quad \\
\cfrac{2x-1-x-3}{x+3}\le 0\)

Answer 16

\(\bf \cfrac{\cancel{ goose }}{\cancel{ goose }}=1\qquad \cfrac{\cancel{ fish }}{\cancel{ fish }}=1\qquad \cfrac{\cancel{ whatever }}{\cancel{ whatever }}=1\) thus

Answer 17

thus \(\bf \cfrac{\cancel{ (x+3) }}{\cancel{ (x+3) }}\to 1\leftarrow \cfrac{(x+3) }{(x+3) }\)

Answer 18

oh okay I see