You can put this solution on YOUR website! xy=12 (equation 1)
x+y=7 (equation 2)
divide equation 1, both sides, by -X --> -y= -12/x --> Y = 12/x (mult by -1)
Substitute Y = 12X into equation 2 for Y
a) X + 12/X =7 (substituting 12X for Y into equation 2)
b) X + 12/X - 7 = 0 (subtract 7 from both sides)
C) X(X +12/X -7) = x(0) (Multiply both sides by X)
X^2 + 12 - 7x = 0 (distribute)
d) (x-3)(X-4) = 0 (factor)
e) Set x-3 =0 and x-4 =0
f) Therefore there are 2 solutions for X : x=3 and x=4
g) Take the first value of X=3 and substitute for X in Equation 2:
h) 3 +y =7 --> Y = 4 Therefore there is an intersection at (3,4)
i) Take the second value of X=4 and substitute into Equation 2
j) 4 + Y =7 ---> Y = 3 Therefore there is an intersection at (4,3)
k) Substitute the 2 solutions back into both equations to make sure there are no extraneous equations.
l) 3+4 =7 : 4+3 =7 (Equation 2)
m) 3*4 =7 : 4*3 =7 (equation 1)
Therefore the solutions are (3,4) and (4,3)
You can put this solution on YOUR website!
Simultaneous equations - one linear, one quadratic
Solve the following equations:
xy = 12
x + y = 7
xy = 12 ---------- eq (i)
x + y = 7_____x = 7 - y -------- eq (ii)
y(7 - y) = 12 ---- Substituting 7 - y for x in eq (i)
(y - 4)(y - 3) = 0
Now substitute each y value into any of the 2 original equations to determine the 2 corresponding values of x.
You'll see that the following solutions are derived:
(
