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To solve:

  1. -3x + 2y = 7
  2. y = x2 - 2x + 3

re-arrange the first to make y the subject and substitute for y in the second:

  1. y = 1/2 (7 + 3x)
  2. 1/2 (7 + 3x) = x2 - 2x + 3

And solving the second:


1/2 (7 + 3x) = x2 - 2x + 3

→ 7 + 3x = 2x2 - 4x + 6

→ 2x2 - 7x - 1 = 0

→ x = 1/4 (7 ± √57)


And substituting back into the (rearranged) first equation:


y = 1/2 (7 + 3x)

= 1/2 (7 + 3(1/4 (7 ± √57)))

= 1/2 (28/4 + 3/4 (7 ± √57)))

= 1/8 (49 ± 3√57)


giving the two solutions:

  1. x = 1/4 (7 + √57) ≈ 3.637, y = 1/8 (49 + 3√57) ≈ 8.956
  2. x = 1/4 (7 - √57) ≈ -0.137, y = 1/8 (49 - 3√57) ≈ 3.294

Note: the value of y is worked out once for each of the +/- of the √57; in this case +√57 for x leads to +√57 for y, and -√57 for x leads to -√57 for y. I showed the calculation in one for both.


To check the result, a substitution can be made in the original equation 2:


x = 1/4 (7 ± √57):

y = (1/4 (7 ± √57))2 - 2(1/4 (7 ± √57)) + 3

= 1/16 (7 ± √57))2 - 8/16(7 ± √57) + 48/16

= 1/16 (49 ± 14√57 + 57 - 56 - ±8√57 + 48)

= 1/16 (96 ± 6√57) [see below]

= 1/8 (49 ± 3√57) as before


In simplifying ±14√57 - ±8√57 = (±14 - ±8)√57 the + and - signs of the 14 and 8 correspond to give:
  • "+14" goes with "- +8" to give 14 - (+8) = 14 - 8 = +6
  • "-14" goes with "- -8" to give -14 - (-8) = -14 + 8 = -6
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Q: How do you algebraically solve these simultaneous equations - 3x plus 2y equals 7 and y equals x2 - 2x plus 3?
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