Any one can help me this question?
1)3y-x=3
2/3y-1/x=2
2)The curve
xy+20=5x and the line 2y=x-3 intersect at points a and b. calculate
the coordinates of a and of b
This kind of questions should be the most basic of the
lot. Like wee_ws has said. Just use the Substitution method.
1)
3y - x = 3
3y = x +
3
- (1)
2/3y - 1/x =
2 -
(2)
Sub (1) into (2)
2/(x+3) - 1/x = 2
[2x - (x+3)]/x(x+3) =
2
{Combine the partial fractions together}
x - 3 = 2x(x+3)
x - 3 = 2x^2 + 6x
2x^2 + 5x + 3 = 0
(x + 1)(2x + 3) = 0
Hence x = -1 & x = -3/2
Sub x = -1 into (1)
3y = -1 + 3
y = 2/3
Sub x = -3/2 into (1)
3y = -3/2 + 3
y = 1/2
Therefore,
x = -1, y = 2/3
x = -3/2, y = 1/2
To check if your values of x & y are correct, substitute
them back into the other equation (2) to see if the LHS = RHS. I'll
leave you to solve the 2nd question on your own. Math is all about
practice so just try it out. Cheers! =)