Either 9−2x is positive or zero or it is negative. If we consider these two possibilities separately we can form equations without modulus signs to find all the possible solutions. We have to be careful, though, because the supposition that 9−2x is positive or negative implies limits on the possible values of x.
Supposing 9−2x is positive or zero x must be less than or equal to 4.5. The equation then becomes 9−2x−3x=10, which gives a solution compatible with the supposition that x≤4.5.
Supposing 9−2x is negative (or zero), x must be greater than or equal to 4.5 for this to be the case. The modulus sign in the equation then converts 9−2x into ⁻(9−2x), which is ⁻9+2x, making the equation ⁻9+2x−3x=10. The solution to this (x=⁻19) is not compatible with the supposition that x≥4.5, so this is not a valid solution to the original equation. Indeed, if you put ⁻19 into the original equation in place of x will will find that it doesn't work.
Since this covers all possibilities the solution to 9−2x−3x=10 (x=⁻0.2) is the only possibility.
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Verified answer
|9 - 2x| - 3x = 10
|9 - 2x| = 10 + 3x
9 - 2x - 3x = 10
9 - 5x = 10
- 5x = 10 - 9
- 5x = 1
x = - 1/5 answer//
|9 – 2x| – 3x = 10 is equivalent to
|9 – 2x| = 3x + 10 (1)
Restriction on x:
|9 – 2x| ≥ 0, therefore
3x + 10 ≥ 0
3x ≥ –10
x ≥ –10/3
Solving (1) by squaring each side
(9 – 2x)² = (3x + 10)²
(9 – 2x)² – (3x + 10)² = 0
(9 – 2x + 3x + 10)(9 – 2x – 3x – 10) = 0
(x + 19)(–5x – 1) = 0
(x + 19)(5x + 1) = 0
Since x ≥ –10/3, x + 19 ≥ –10/3 + 19 > 0,
therefore x + 19 ≠ 0. Hence
5x + 1 = 0
x = –1/5
(–1/5 > -10/3)
Answer: –1/5 is the sole solution to the given equation.
Either 9−2x is positive or zero or it is negative. If we consider these two possibilities separately we can form equations without modulus signs to find all the possible solutions. We have to be careful, though, because the supposition that 9−2x is positive or negative implies limits on the possible values of x.
Supposing 9−2x is positive or zero x must be less than or equal to 4.5. The equation then becomes 9−2x−3x=10, which gives a solution compatible with the supposition that x≤4.5.
Supposing 9−2x is negative (or zero), x must be greater than or equal to 4.5 for this to be the case. The modulus sign in the equation then converts 9−2x into ⁻(9−2x), which is ⁻9+2x, making the equation ⁻9+2x−3x=10. The solution to this (x=⁻19) is not compatible with the supposition that x≥4.5, so this is not a valid solution to the original equation. Indeed, if you put ⁻19 into the original equation in place of x will will find that it doesn't work.
Since this covers all possibilities the solution to 9−2x−3x=10 (x=⁻0.2) is the only possibility.
There are two solutions because the first part is an indefinite part, so it can end like this:
|9-2x|-3x=10
9-2x-3x=10
-5x=1
x=-1/5
or it can end like this
-|9-2x|-3x=10
-9+2x-3x=10
x=19
|9-2x|-3x=10
|9-2x|=3x+10
Two equations
1] 9-2x=3x+10; 5x=-1; x=-1/5
2] 2x-9=3x+10; x=-19
Now try both solutions
|9- 2(-1/5)| = 3(-1/5)+10
9+2/5= 10-3/5
47/5=47/5; solution checks
|9-2(-19)|= 10+3(-19)
47= -47; solution void
Solution is x=-1/5