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Dillesh4096
If lx + 3l = 4x - 5. Find the value(s) of x.

A. 2/5
B. 8/3
C. 8/5
D. 2/5, 8/3
E. None

Solving
lx + 3l = 4x - 5

Case I:
when x+3>0

x+3=4x-5
3x=8
x=8/3

Case2: x+3<0

x+3=5-4x
5x=2
x=2/5

So ans: x=2/5,8/3

plugin the values back in eqn
only 8/3 satisfies the eqn
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lx + 3l = 4x - 5

=> x + 3 = 4x - 5 and x + 3 = - (4x - 5)

=> 8 = 3x and 5x = 2

=> x = \(\frac{8}{3}\) and x = \(\frac{2}{5}\)

Answer B
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but what is the reason that 2/5 is not satisfying the equation.
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MathRevolution
lx + 3l = 4x - 5

=> x + 3 = 4x - 5 and x + 3 = - (4x - 5)

=> 8 = 3x and 5x = 2

=> x = \(\frac{8}{3}\) and x = \(\frac{2}{5}\)

Answer B

lets say you solve this quadratic equation: x^2-4x-5=0 to get the values of x as x=-1 and x=5. will you put these value back to cross verify whether the roots you have found out are correct or just mark the answer. I think you ll go with the later one.
Then why are we putting the values back in this question. And even if we are putting back why is that the value 2/5 is not satisfying the equation.
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verma8670
but what is the reason that 2/5 is not satisfying the equation.

Recall that for the absolute value function, it only admits non-negative X. If you are in a situation in which you have x on both sides of an equation involving the absolute value function, it is important that you check your answers - because you might have a choice that leads to a negative x value, which is not possible.
In this case, if you plug in x=2/5 on the RHS, you will find that it becomes 8/5 less 5 on the RHS. This is clearly a negative number, and thus it is not admitted as a value of X.
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|x + 3| = 4x - 5

As we have |x +3| in the equation so we will have two cases
-Case 1: x + 3 ≥ 0 => x ≥ -3

=> |x + 3| = x + 3
=> x + 3 = 4x - 5
=> 3x = 8
=> x = 8/3

But condition was x ≥ -3 and 8/3 ≥ -3
=> x = 8/3 is a SOLUTION
-Case 2: x + 3 ≤ 0 => x ≤ - 3

=> |x + 3| = -(x + 3)
=> -(x + 3) = 4x - 5
=> 5x = 2
=> x = 2/5

But condition was x ≤ 0 and 2/5 is NOT ≤ 0
=> NO SOLUTION

So, Answer will be B
Hope it helps!

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