Given: f(x) = x + 2 and h(x) = 1/ x - 1
2f(x) - h(x) = ?
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OpenStudy (eliesaab):
2 f(x)= 2( x+2)=?
jhonyy9 (jhonyy9):
h(x) = (1/x) -1 or h(x) = 1/(x-1) ???
OpenStudy (red3134):
@jhonyy9 h(x) = 1/(x-1)
jhonyy9 (jhonyy9):
ok. - ty
so can you calculi what wrote above eliesaab ?
OpenStudy (red3134):
@jhonyy9 Not really...
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jhonyy9 (jhonyy9):
this will give you 2f(x) = ?
jhonyy9 (jhonyy9):
why ?
f(x) = x+2 and this you need multiply it by 2 so will be 2(x+2) = ?
jhonyy9 (jhonyy9):
how you distribute this parentheses ?
OpenStudy (red3134):
@jhonyy9 would it be 2x + 4?
jhonyy9 (jhonyy9):
exactly - very nice
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jhonyy9 (jhonyy9):
so and now 2f(x) -h(x) = ?
OpenStudy (red3134):
@jhonyy9 (2x + 4) - 1/ (x - 1) ?
jhonyy9 (jhonyy9):
yes and make the calculi
jhonyy9 (jhonyy9):
a-1/b = ?
jhonyy9 (jhonyy9):
how you subtract from one integer a fraction ?
a-(1/b) = ?
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jhonyy9 (jhonyy9):
how many will be the common denominator ?
jhonyy9 (jhonyy9):
in case of my example the common denominator will be b so this mean that you need multiply the a by b and will get (a*b -1)/b
ok. ? can use this example to solve your exercise ?
jhonyy9 (jhonyy9):
what is b in case of your exercise ?
OpenStudy (red3134):
jhonyy9 (jhonyy9):
please use this example - so b in case of your exercise will be (x-1)
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OpenStudy (red3134):
Would a be 2x + 4?
OpenStudy (red3134):
@jhonyy9 And 2x + 4 multiplied by x - 1 would be 2x - 4?
jhonyy9 (jhonyy9):
no
2x+4 - ( 1/(x-1) ) = ?
jhonyy9 (jhonyy9):
(2x+4)(x-1) = ?
jhonyy9 (jhonyy9):
= 2x*x +2x*(-1) +4*x +4*(-1) = ?
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OpenStudy (red3134):
\[2x ^{2}+ 2x - 4\]
jhonyy9 (jhonyy9):
yes but dont forget the denominator
2f(x) -h(x) = ?
jhonyy9 (jhonyy9):
looke please
in case of my example the common denominator will be b so this mean that you need multiply the a by b and will get (a*b -1)/b
ok. ? can use this example to solve your exercise ?
jhonyy9 (jhonyy9):
this what you ve got above is just the part of a*b
jhonyy9 (jhonyy9):
do you understand it now ?
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jhonyy9 (jhonyy9):
but you need getting (a*b -1)/b
jhonyy9 (jhonyy9):
so will be
2f(x) -h(x) = 2x+4 - (1/(x-1)) = [(2x+4)(x-1)-1]/(x-1) = ?
jhonyy9 (jhonyy9):
substitute there in place of (2x+4)(x-1) this above got value
jhonyy9 (jhonyy9):
= (2x^2 +2x-4 -1)/(x-1) = ?
jhonyy9 (jhonyy9):
-4-1 = ?
this is the last step
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