Consider the linear equations a+b=2 and 2a+b=6. Elimination by subtraction gives a=4 and b=-2. Substitution of b=2-a into 2a+b=6 gives 2a+2-a=6, a=4 and b=-2. No problem either method works.
Now consider 2a+b=6 and 4a+2b=9. Double the first equation: 4a+2b=12. Eliminate by subtraction: 0=3, which is false, so elimination failed. Try substitution: b=6-2a; 4a+2(6-2a)=9; 4a+12-4a=9; 12=9. Substitution fails, too, (as expected).
Next consider 2a+b=6 and 4a+2b=12. Elimination and substitution will give you 0=0, which is true, but will not help you find unique values for a and b because there are many solutions. All you have is a relationship between a and b.
So FALSE, because failure of elimination (if correctly applied) implies failure of substitution.
(The situation is different for non-linear systems.)
Next consider 2a+b=7 and a=4. Elimination can still be carried out by doubling the second equation: 2a=8 and subtracting: b=-1. And substitution gives the same result (as expected).
Therefore, if elimination fails, substitution will also fail in a linear system, and there is no point in trying it.