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Elanso [62]
3 years ago
15

How the heck do you solve these type of problems?

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
6 0
Do the math the same way you would if all the other variables were numbers. For linear equations such as these, you can
• subtract one side of the equation, so you get an equation equal to zero.
• factor out the variable of interest
• subtract all terms that don't have the variable as a factor
• divide by the coefficient of the variable.

19. 2xy - x - 7 = 0 . . . . subract the right side
  x(2y-1) -7 = 0 . . . . factor out x
  x(2y -1) = 7 . . . . subtract the term(s) not containing x. Subtract -7 here.
  x = 7/(2y -1) . . . . divide by the coefficient of x

20. Usually for these, you would eliminate fractions first. (You don't have to, but it does simplify things a bit.)
  2d + f = 24 . . . . multiply the equation by 4 to eliminate fractions
  f = 24 - 2d . . . . subtract the term not involving f

21. We can go back to the above sequence of steps here, if you like.
  3(2j -k) -108 = 0 . . . . subtract the right side
  6j -3k -108 = 0 . . . . separate the j terms from the others (eliminate parentheses using the distributive property)
  6j = 108 + 3k . . . . add the opposite of the non-j terms
  j = 108/6 + 3k/6 . . . . divide by the coefficient of j
  j = 18 + k/2 . . . . simplify

22. "solve for" variable not specified.
  t = (3.5s -42)/4
  s = (4t+42)/3.5
In both cases, the 3.5 can be replaced by 7/2 and the fraction simplified.
  t = (7/8)(s-12)
  s = 12 + (8/7)t
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