Question

Can someone please explain to me how you solve inequalities with a fraction on one side. For example: x/8+3<15. I do not know how to do this. By the way it is in Algebra II.

Answer 1

more or less same as you'd any linear equation, how would you solve say \(\bf \cfrac{x}{8}+3=15\)

Answer 2

I thought you would multiply x/8 by 8 to get rid of the denominator and then multiply 15 by 8 but I am not for sure.

Answer 3

\(\bf \cfrac{x}{8}+3<15\\
\textit{subtracting 3 from both sides}\\
\cfrac{x}{8}\cancel{+3-3}<15-3 \implies \cfrac{x}{8}<12\\
\textit{multiplying 8 to both sides}\\
\cfrac{x}{\cancel{8}}\times \cancel{8}<12 \times 8 \)

Answer 4

Is this how you do all of them that have a fraction on one side even if the fraction is negative?

Answer 5

nope, that's the main difference in an inequality, if you multiply/divide/exponentialize both sides by a NEGATIVE value, then you'd need to \(\bf \text{flip the sign}\)

lemme redo that for the sake of example, using a negative fraction
\(\bf -\cfrac{x}{8}+3\color{blue}{<}15\\
\textit{subtracting 3 from both sides}\\
-\cfrac{x}{8}+3-3<15-3 \implies -\cfrac{x}{8}<12\\
\textit{multiplying -8 to both sides}\\
\cfrac{x}{8}\times -8<12 \times -8 \implies x \color{blue}{>} -96\)

Answer 6

woops, I missed the -... anyhow \(\bf -\cfrac{x}{8}\times -8<12 \times -8 \implies x \color{blue}{>} -96\)

Answer 7

but you would still subtract/add the whole number multiply by the denominator and then if its negative then flip the sign

Answer 8

the procedure is the same as in any equality, the difference is when the multiplier/divider/exponent is negative, otherwise yes

Answer 9

Okay and what about -1/4x-1<=-9

Answer 10

same thing

Answer 11

so itd be +1 and then multiply both sides by -1/4 and thatll give you x=> 2 right

Answer 12

\(\bf \cfrac{1}{4x}-1\le -9\\
\textit{adding 1 to both sides }\\
\cfrac{1}{4x}\cancel{-1+1}\le -9+1 \implies \cfrac{1}{4x}\le -8\\
\textit{multiplying both sides by 4x}\\
\cfrac{1}{\cancel{4x}}(\cancel{4x})\le -8(4x)\)

Answer 13

the 1/4 is negative

Answer 14

ohhh... ok

Answer 15

yeah

Answer 16

\(\bf -\cfrac{1}{4x}-1\le -9\\
\textit{adding 1 to both sides }\\
-\cfrac{1}{4x}\cancel{-1+1}\le -9+1 \implies -\cfrac{1}{4x}\le -8\\
\textit{multiplying both sides by -4x}\\
-\cfrac{1}{\cancel{4x}}(\cancel{-4x})\le -8(-4x)\)

Answer 17

Im still not getting the correct answer its coming out to -1/32 or -0.031

Answer 18

so that'd leave us with \(\bf 1 \ge 32x \implies \cfrac{1}{32} \ge x\)

notice, on the -4x, negative, the sign flpped

Answer 19

well you see \(\bf -\cfrac{1}{4x}(-4x)\le -8(-4x) \implies \cfrac{1}{-4x}(-4x)\le -8(-4x)\)

Answer 20

Im not getting a whole number

Answer 21

yeap, doesn't have to be

Answer 22

my answers on the work sheet are x>-80, x>96, x>21, x>-54, x=>32, x=>-6, or x =>-91. I am doing a riddle work sheet called "What Happened When the King of Beasts Runs in Front of a Train?"

Answer 23

well, that just means your inequality is incorrect

Answer 24

Oh I dont know I will just have to ask my teacher then