Question

Solve for x, y and z in the below Linear equation by substitution method :
$10x+5y-8z=-55x-4y+z=3x+2y-z=3$

• A.
  $2,4and5$
• B.
  $2,3and5$
• C.
  $1,2and4$
• D.
  $0,4and8$

Answer 1

The correct option is B

$2,3and5$
We have,
$10x+5y-8z=-\mathrm{5...}\left(i\right)5x-4y+z=\mathrm{3...}\left(ii\right)x+2y-z=\mathrm{3...}\left(iii\right)$

Multiplying (ii) by 2 and subtracting from (i), we get,
$10x+5y-8z-10x+8y-2z=-5-6\Rightarrow 13y-10z=-\mathrm{11...}\left(iv\right)$

Multiplying (iii) by 5 and subtracting from (ii), we get,

$5x-4y+z-5x-10y+5z=3-15\Rightarrow -14y+6z=-12\Rightarrow 7y-3z=\mathrm{6..}\left(v\right)$

Multiplying (v) with 10 and subtracting from 3 times (iv), we get,

$39y-30z-70y+30z=-33-60\Rightarrow -31y=-93\Rightarrow y=3From\left(iv\right)39-10z=-11\Rightarrow z=5From\left(iii\right)x+6-5=3\Rightarrow x=2$

So, option b is correct.