Solving this absolute value equation $|3(x-1)-10| = |5(x+2)-3x|$

Question

If $|3(x-1)-10| = |5(x+2)-3x|$, i.e., $|3x-13| = |2x +10|$, then we have 2 linear equations: $$3x-13 = 2x+10, \text{ or }x = 23,$$

and

$$3x-13 = -2x-10, \text{ or } x = \frac{3}{5}.$$

Are those the only solutions to the equation?

Answer 1

Are those the only solutions to the equation?

Yes. Your approach is totally correct. There is only real solution possible.

Let's check with graph:

Answer 2

$|3x-13|=|2x+10|$

Put $x=23,$

LHS=$|3×23-13|=|69-13|=56$

RHS=$|2×23+10|=|46+10|=56$

Again,put $x=\frac 35$

LHS=$|3×\frac 35-13|=|\frac{9}5-13|=|\frac{9-65}5|=|\frac{56}5|=\frac{56}5$

RHS=$|2×\frac35+10|$=$|\frac 65+10|$=$\frac{56}5$

So,both are the solutions,since both of them satisfy the equation.

Answer 3

$|3x-13|=|2x+10|$

squaring on both sides

$$9x^2+169-78x=4x^2+100+40x$$ $$5x^2-118x+69=0$$ $$x=\frac{118 \pm \sqrt{118^2-4\cdot 5 \cdot69}}{10}$$ $$x=\frac{118 \pm 112 }{10}$$ $$x=\frac{3}{5}\text{ or } x=23 $$