Question

solve (x-3)/(x-4)>equalto 2

Answer 1

\[\frac{x-3}{x-4}-2 \ge 0\]
first step I would do is combine fractions on the left

Answer 2

do u need answer or the solution @IloveHW

Answer 3

then find when that fraction is zero and undefined
then draw a number line
then test the intervals around each number that gave us the fraction was undefined or zero

Answer 4

-3 = -4+1 just a stray thought tho

Answer 5

x belongs to (4,5]

Answer 6

1 + 1/(x-4) >= 2

1/(x-4) >= 1

Answer 7

in order for this to be true, the bottom has to be positive soo the sign can stay the same direction and

1 >= x-4

Answer 8

can someone take me step by step i dont just want the answer

Answer 9

tada!!

Answer 10

ohk @IloveHW

Answer 11

multiply both sides by the square of the denominator

\[\frac{(x - 3)(x - 4)^2}{(x - 4)} \ge 2(x - 4)^2\]

remembering

\[x \neq 4\]

then solve for x

Answer 12

its not square btw

Answer 13

this is wrong @campbell_st never do it like this

Answer 14

wrong process..
let me help u out @IloveHW

Answer 15

(x-4)^2 is positive
I don't see why he can't do it that way

Answer 16

he multiplied both sides by a positive number which doesn't effect the direction of the inequality

Answer 17

I know, you take the denominator and square it.... then multiply both sides of the inequality by that

Answer 18

it gives you

\[x^2 - 7x + 12 \ge 2x^2 - 16x + 32\]

next collect like terms... etc

Answer 19

the solution is x|4<x<eqto5

Answer 20

\[\frac{x-3}{x-4} -2 \ge 0 \\ \frac{x-3}{x-4}-\frac{2(x-4)}{x-4} \ge 0 \\ \frac{x-3-2(x-4)}{x-4} \ge 0\]
find when bottom is zero and when top is zero
then put those numbers on a number line
and test the intervals around those numbers to see which are positive or not