Or, Subtract eq2 from eq1 Get x^2 - y^2 = 7 (x+y)(x-y) = 7 Assume x+y=7 and x-y=1 Add those two new equalities Get 2x=8, x=4 x+y=7 => y=3 xy=12 Edit: the assumtion part is based on 7 being a prime, and 1 and 7 being its factors, therefore making the parenthesis with addition 7 and the one with subtraction 1. This works because for x and y being integers this is the obvious solution, and people making problems like these like making whole number solutions
There's no need to assume anything. Including that x & y are integers. Add both equations, then subtract the 2nd from the first. These operations yield x² + 2xy + y² = 49; x+y = ±7 and x² - y² = 7; (x+y)(x-y) = 7 Combine both results to get x-y = ±1 Knowing their sum & difference, x & y are easily found. x+y = 7, x-y = 1; (x,y) = (4,3); xy = 12 or x+y = -7, x-y = -1; (x,y) = (-4,-3); xy = 12 In both cases, xy = 12 Fred
Or,
Subtract eq2 from eq1
Get x^2 - y^2 = 7
(x+y)(x-y) = 7
Assume x+y=7 and x-y=1
Add those two new equalities
Get 2x=8, x=4
x+y=7 => y=3
xy=12
Edit: the assumtion part is based on 7 being a prime, and 1 and 7 being its factors, therefore making the parenthesis with addition 7 and the one with subtraction 1. This works because for x and y being integers this is the obvious solution, and people making problems like these like making whole number solutions
@@JTtheking134 Excellent
There's no need to assume anything. Including that x & y are integers.
Add both equations, then subtract the 2nd from the first. These operations yield
x² + 2xy + y² = 49; x+y = ±7
and
x² - y² = 7; (x+y)(x-y) = 7
Combine both results to get
x-y = ±1
Knowing their sum & difference, x & y are easily found.
x+y = 7, x-y = 1; (x,y) = (4,3); xy = 12
or
x+y = -7, x-y = -1; (x,y) = (-4,-3); xy = 12
In both cases, xy = 12
Fred
which year olympiad is this(e.g., yr 8,9,10 etc/ 8th, 9th, 10th etc)?
x=4 and y=3.
@@adgf1x the value of x and y is not only 4 and 3 also -4 and -3
XYZ
@@ChinSoonLeow 🙄