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5. What is the value of integer x?
(1) 2x^2+9<9x
(2) |x+10|=2x+8
Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039650
Please have a look at Bunuel's solution for details. I have 2 specific question:
1. when 2x^2+9<9x is solved it comes to (2x-3)(x-3)<0 Implies (please poke a hole in my either or logic as the solution provided is confusing me a bit currently
either x-3 <0 implies x< 3
or 2x - 3 < 0 implies x< 3/2
Now if the above is drawn on a number line the values of X would look like
-5-4-3 -2 -1 0 1 2 3 4 5
Left of +3
or
Left of 3/2
In essence implies X (integer) could be lie anywhere left of 3 ? Rather than X lies between 1.5 and 3 leading to the solution of 2 and thus sufficient? thats 2x-3 > 0 - HRS vs LHS makes a difference?
Happy to listen to alternative suggestions
2. |x+10|=2x+8
For the above situations there could be 2 situations either of = or > <
Bunuels solution by logic that the LHS is always positive and thus solve RHS first sounds a bit unintuitive in an algebraic equation. Generally speaking algebraic solutions are supposed to be solved together ?
to keep all sides positive we could square both sides and then solve
or alternatively
do the +ve and -ve for x+10 and solve?
But first solving HRS and then replacing that solution - should this be the way if a combination of equality and equation comes across?
What if
|x+10| = |2x+8|
What should be done then
or
|x+10| > 2x+8 or equivalent?
Please suggest
(1) 2x^2+9<9x
(2) |x+10|=2x+8
Solution: the-discreet-charm-of-the-ds-126962-40.html#p1039650
Please have a look at Bunuel's solution for details. I have 2 specific question:
1. when 2x^2+9<9x is solved it comes to (2x-3)(x-3)<0 Implies (please poke a hole in my either or logic as the solution provided is confusing me a bit currently
either x-3 <0 implies x< 3
or 2x - 3 < 0 implies x< 3/2
Now if the above is drawn on a number line the values of X would look like
-5-4-3 -2 -1 0 1 2 3 4 5
Left of +3
or
Left of 3/2
In essence implies X (integer) could be lie anywhere left of 3 ? Rather than X lies between 1.5 and 3 leading to the solution of 2 and thus sufficient? thats 2x-3 > 0 - HRS vs LHS makes a difference?
Happy to listen to alternative suggestions
2. |x+10|=2x+8
For the above situations there could be 2 situations either of = or > <
Bunuels solution by logic that the LHS is always positive and thus solve RHS first sounds a bit unintuitive in an algebraic equation. Generally speaking algebraic solutions are supposed to be solved together ?
to keep all sides positive we could square both sides and then solve
or alternatively
do the +ve and -ve for x+10 and solve?
But first solving HRS and then replacing that solution - should this be the way if a combination of equality and equation comes across?
What if
|x+10| = |2x+8|
What should be done then
or
|x+10| > 2x+8 or equivalent?
Please suggest
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