anonymous
  • anonymous
If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation? A. y, in the first equation B. x, in the second equation C. x, in the first equation D. y, in the second equation
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
click on it

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@johnweldon1993
triciaal
  • triciaal
the second can be factored by 3 then use x = expression to be substituted in the first
johnweldon1993
  • johnweldon1993
Hmm, I would solve for 'x' in the second equation...by why? Notice how in the equation \(\large 3x -12y = 15\) every number is divisible by 3
johnweldon1993
  • johnweldon1993
If we were to solve that for 'x' we have [\large x = 5 + 4y\] nice an even numbers... whereas in the first equation \(\large 2x + 6y = 9\) no matter what we solve for ...not everything is divisible by 2...or 6 ...so that would give s fractions...and who wants that lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.