Step-by-step explanation:
or,(x-2)(x-4)+(3x-11)(x-3)/(x-3)(x-4)=(4x+13)/(x+1)
or,x^2-4x-2x+8+3x^2-9x-11x+33/x^2-4x-3x+12=(4x+13)/(x+1)
or,4x^2-4x-2x-9x-11x+8+33/x^2-7x+12=(4x+13)/(x+1)
or,4x^2-26x+41/x^2-7x+12=(4x+13)/(x+1)
or, (x+1)(4x^2-26x+41)=(4x+13)(x^2-7x+12)
or,4x^3-26x^2+41x+4x^2-26x+41=4x^3-28x^2+48x+13x^2-91x+156
or,-26x^2+4x^2+28x^2-13x^2+41x-26x-48x+91x=156-41
or,-7x^2+58x=115
or,-7x^2+58x-115=0
or,-(7x^2-58x+115)=0
or,7x^2-58x+115=0
or,7x^2-(35+23)x+115=0
or,7x^2-35x-23x+115=0
or,7x (x-5)-23 (x-5)=0
or, (x-5)(7x-23)=0
Either, Or,
(x-5)=0 (7x-23)=0
or,x=0+5 or,7x=23
:x=5 :x=23/7
Hence,x=5 or 23/7
Answer:
x+11
Step-by-step explanation:
here is the equation
PART 1:
Jeremy gives the correct answer.
The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....
Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value.
PART 2:
The long division method is shown in the picture below
PART 3:
As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct