Question 712600
Given
(1) x/2 - y/3 = 5/6 and
(2) x/5 - y/4 = 57/10
Clear the fractions in (1) by multiplying both sides of the equation by 6 and get
(3) 3x - 2y = 5
Clear the fractions in (2) by multiplying both sides of the equation by 20 and get
(4) 4x - 5y = 114
Now solve (3) for x and get
(5) 3x = 2y + 5 or
(6) x = (2/3)y + 5/3
Now substitute x of (6) into (4) and get
(7) 4*((2/3)y + 5/3) - 5y = 114 or
(8) (8/3)y + 20/3 - 5y = 114 or
(9) (8/3 - 5)y = 114 - 20/3 or
(10) ((8-15)/3)y = (342-20)/3 or
(11) -(7/3)y = 322/3 or
(12) -7y = 322 or
(13) y = -322/7 or
(13) y = -46
Use y of (13) in (6) to get x
(14) x = (2/3)(-46) + 5/3 or
(15) x = (-92 + 5)/3 or
(16) x = -87/3 or
(17) x = -29
Check this pair with (2).
Is (-29/5 -(-46)/4 = 57/10)?
Is ((-29*4 + 46*5)/20 = 57/10)?
Is ((-116 + 230)/20 = 57/10)?
Is (114/20 = 57/10)?
Is (57/10 = 57/10)? Yes
Answer: The solution pair, using fraction clearing and substitution, is
 (-29,-46)