Answer:
I'll choose 'x' variable first to eliminate because -
- -3 (coefficient of 'x' in Eqn.1) & 3 (coefficient of 'x' in Eqn.2) are additive inverses of each other , so they can be easily eliminated by simply adding those equations.
- It can be also solved by using elimination method but you'll need to multiply -1 with either Eqn.1 or Eqn.2 because in elimination method , a variable can be eliminated only when their coefficients are numerically equal (along with their sign).
Step-by-step explanation:
Eqn.1 → -3x - 8y = -24
Eqn.2 → 3x - 5y = 45
If we'll add both the equations , then '3x' would be getting cancelled first.
⇒ [tex](-3x - 8y) + (3x - 5y) = (-24) + (45)[/tex]
⇒ [tex]3x - 3x - 8y - 5y = 45 - 24[/tex]
⇒ [tex]-13y = 21[/tex]
⇒ [tex]y = \frac{-21}{13}[/tex]
Putting the value of y in eqn.2 ,
⇒ [tex]3x - 5(\frac{-21}{13}) = 45[/tex]
⇒ [tex]3x + \frac{105}{13} = 45[/tex]
⇒ [tex]3x = 45 - \frac{105}{13}[/tex]
⇒ [tex]3x = \frac{585 - 105}{13} = \frac{480}{13}[/tex]
⇒ [tex]x = \frac{480}{13 \times 3} = \frac{480}{39}[/tex]
Reducing the fraction gives ,
⇒ [tex]x = \frac{160}{13}[/tex]
