Find the points of intersection of a straight line 2x +3y = 10 and the curve 2/x + 3/y = 5.
Answer: (4, 2/3)(1/2, 3)
Given that (-2, 2) is a solution to the simultaneous equations,
x + 1/2y = h/2 and 1/x + 2/y = k,
find the value of h and of k. Hence find the other solution.
Answer: h = -2, k = 1/2, (1, -4)
Can someone help me out with those 2 qns.
Thanks in advance.
"Find the points of intersection of a straight line 2x +3y = 10 and the curve 2/x + 3/y = 5.
Answer: (4, 2/3)(1/2, 3)"
2x + 3y = 10 ----- ( 1 )
2/x + 3/y = 5 ----- ( 2 )
From ( 1 )
2x = 10 - 3y
x = (10-3y)/2 ----- ( 3 ) <<<< can either use x or y as subject.
Sub ( 3 ) into ( 2 )
2/((10-3y)/2) + 3/y = 5
4/(10-3y) + 3/y = 5
(4y + 30 - 9y)/(10y - 3y² ) = 5
4y + 30 - 9y = 50y - 15y²
15y² - 55y + 30 = 0
3y² - 11y + 6 = 0
(3y - 2)(y - 3) = 0
(3y - 2) = 0 or (y - 3) = 0
y = 2/3, 3
Sub in Y values and get x
When y = 2/3, x = 4
When y = 3, x = 1/2
"Given that (-2, 2) is a solution to the simultaneous equations,
x + 1/2y = h/2 and 1/x + 2/y = k,
find the value of h and of k. Hence find the other solution.
Answer: h = -2, k = 1/2, (1, -4)"
x + 1/2 y = h/2 ----- ( 1 )
1/x + 2/y ----- ( 2 )
Sub (-2,2) into ( 1 )
-2 + 1 = h/2
h = -2
Sub (-2,2) into ( 2 )
1/-2 + 2/2 = k
k = 1/2
Sub k = 1/2 and h = -2 into ( 1 ) and ( 2 )
Now,
x + 1/2y = -1 ----- ( 1 )
1/x + 2/y = 1/2 ----- ( 2 )
From ( 1 )
x = -y/2 - 1 ----- ( 3 )
Sub ( 3 ) into ( 2 )
1/(-y/2 - 1) + 2/y = 1/2
(y + 2(-y/2 - 1)/(y( -y/2 - 1) = 1/2
-2/(y²/2 - y ) = 1/2
-4 = -y²/2 - y
y² + 2y -8 = 0
( y + 4)( y - 2) = 0
y + 4 = 0 or y - 2 = 0
y = 2( 1st solution)
y = -4 ( 2nd solution)
Sub y = -4 into ( 3 )
x = 1
Hence, the other solution is ( 1, -4)