An algebra problem by nikhil jaiswal
How many solutions are there to
∣ x + 2 ∣ + ∣ x − 5 ∣ = 7 ?
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4 solutions
i'd derived the same closed interval but i made the mistake of counting the integer solutions and went for 8 not even checking the fractional values...lol
given equation should have 4 solution.. I think
there are infinetely many solutions to this equation
how to prove it?
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try 2.1. that will add one more solution, making it none of these
Just 2 solutions .... It will be made with guessing At first time we will guess both of them are +ve .. then the answer will be :
x+2 + x-5 =7; 2x-3=7; 2x = 10 => x=5
At the second time we will guess both of them are -ve ... then the answer will be :
-x-2-x+5 = 7; -2x + 3 = 7; -2x = 4 => x = -2
When we guess The first abs is +ve and the second is -ve the equation will be 7 = 7 A useless equation abd the first abs is -ve and the second is +ve the eqn. will be -7 = 7 A wrong eqn.
so the answer is... 2 solutions
But your 'useless equation' tells you that there are in fact an infinite number of solutions!
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How ? "infinite number of sln." and no variables?
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There are variables - they cancel out!
If we assume x + 2 is positive and x − 5 is negative, then the equation becomes x + 2 − x + 5 = 7 . Whatever value of x we use, this simplifies to 7 = 7 - so all values of x (an infinite number of them!) solve this equation!
Of course, our initial assumption is only true for − 2 < x < 5 , but this still leaves an infinite number of solutions.
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@J Thompson – Well try and substitute any value for x like you say and see if it works. The case here is that For whatever value of X you get the other modulus function cant get a negative of that X so both Xs must have the same sign basically. So they are either both positive or both negative.
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@Farouk Yasser – We're not saying that the x values have different signs, we're saying that the expressions inside the modulus functions have different signs.
If x = 1 , then x + 2 = 3 (which is positive), and x − 5 = − 4 (which is negative).
What the modulus function does is turn negative values into positive ones. Still using x = 1 : ∣ x + 2 ∣ = ∣ 3 ∣ = 3 (positive) and ∣ x − 5 ∣ = ∣ − 4 ∣ = 4 (positive).
The sum of these two values is obviously 7 .
As I said above, this will work for any x ∈ [ − 2 , 5 ] .
@J Thompson – you are right but 7=7 means there exist infinitly many solvs
how come there are 8 solution when I get -2 and 5 as answers???
this is the best solution, i mean the 1st solution
The answer is 2 solutions ie., -2 and 5 ==> None of these for the given options.
what about all other x ∈ R ∣ 2 ⩽ x ⩽ 5 ? infinite solutions
For x ≤ − 2 we have
∣ x + 2 ∣ = − ( x + 2 ) = − x − 2 and ∣ x − 5 ∣ = − ( x − 5 ) = − x + 5 .
Thus for x ≤ − 2 we have that
∣ x + 2 ∣ + ∣ x − 5 ∣ = 7 ⟹ − x − 2 − x + 5 = 7
⟹ − 2 x = 4 ⟹ x = − 2 ,
which lies on the interval x ≤ − 2 and hence is a valid solution.
For − 2 ≤ x ≤ 5 we have
∣ x + 2 ∣ = x + 2 and ∣ x − 5 ∣ = − ( x − 5 ) = − x + 5 .
Thus for − 2 ≤ x ≤ 5 we have that
∣ x + 2 ∣ + ∣ x − 5 ∣ = 7 ⟹ x + 2 − x + 5 = 7 ⟹ 7 = 7 ,
which is a tautology and hence valid for all x on this interval, i.e. there are an infinite number of solutions on this interval.
For x ≥ 5 we have that
∣ x + 2 ∣ = x + 2 and ∣ x − 5 ∣ = x − 5 .
Thus for x ≥ 5 we have that
∣ x + 2 ∣ + ∣ x − 5 ∣ = 7 ⟹ x + 2 + x − 5 = 7 ⟹ 2 x = 1 0 ⟹ x = 5 ,
which lies on the interval x ≥ 5 and hence is a valid solution.
Thus the solution set is the interval [ − 2 , 5 ] , i.e., an infinite number of solutions, making "none of these" the correct option.