Find hyperboles y=-5/x and the line 7,5x +3y -7,5=0 cutting points.
First I did this:
7,5x +3y -7,5=0
3y=-7,5x +7,5 | /3
y=-2,5x +2,5 (To get it into y=ax +bx form)
Now I know that y equals -2,5x +2,5 so I put it into the equation:
-2,5x +2,5=-5/x
-2,5x +x=-2,5 -5
-1,5x=-7,5 | /(-1,5)
x=5
Replacing x in y=-5/x
y=-5/5
y=-1
First pair of cutting points is (5;-1), which I checked is correct.
Now how do I find the other pair, I've tried in every possible way, but no prevail.
How do I get the other pair? Other pair should be: (2;-2,5)
Thanks!
First I did this:
7,5x +3y -7,5=0
3y=-7,5x +7,5 | /3
y=-2,5x +2,5 (To get it into y=ax +bx form)
Now I know that y equals -2,5x +2,5 so I put it into the equation:
-2,5x +2,5=-5/x
-2,5x +x=-2,5 -5
-1,5x=-7,5 | /(-1,5)
x=5
Replacing x in y=-5/x
y=-5/5
y=-1
First pair of cutting points is (5;-1), which I checked is correct.
Now how do I find the other pair, I've tried in every possible way, but no prevail.
How do I get the other pair? Other pair should be: (2;-2,5)
Thanks!
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y = - 5 / x <==
7.5x + 3y - 7.5 = 0
3y = - 7.5x + 7.5
y = - 2.5x + 2.5 <==
- 5 / x = - 2.5x + 2.5
- 2.5x^2 + 2.5x + 5 = 0
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1) * (x - 2) = 0
x = -1 ; x = 2
y = - 5 / -1 ; -5 / 2
y = 5 ; y = - 2.5
( -1 , 5) and (2 , - 2.5)
7.5x + 3y - 7.5 = 0
3y = - 7.5x + 7.5
y = - 2.5x + 2.5 <==
- 5 / x = - 2.5x + 2.5
- 2.5x^2 + 2.5x + 5 = 0
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1) * (x - 2) = 0
x = -1 ; x = 2
y = - 5 / -1 ; -5 / 2
y = 5 ; y = - 2.5
( -1 , 5) and (2 , - 2.5)
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You're welcome, and aware that some use a comma.
I try not to pay much attention to spelling, grammar, or typos...
This is math, not language skills.
That said, how would you know of a mistake unless it was pointed out?
I didn't think either meant to be rude or condescending.
I try not to pay much attention to spelling, grammar, or typos...
This is math, not language skills.
That said, how would you know of a mistake unless it was pointed out?
I didn't think either meant to be rude or condescending.
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...both sets of answers should come from the substitution...you have mistakes...
solving the linear equation for y: y = (5/2) - (5x/2) = (5 - 5x)/2
substituting...
(5 - 5x)/2 = - 5/x
this is a quadratic...
5x - 5x^2 = - 10
5x^2 - 5x - 10 = 0
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
there are your two values for x...y values follow easily...
[Your first solution---(5, -1)---is not correct...the coordinate pair is (- 1, 5)]
Not to make this too dramatic, or to "overstate the case", the word you want is:
HYPERBOLA
[look up hyperbole !]
solving the linear equation for y: y = (5/2) - (5x/2) = (5 - 5x)/2
substituting...
(5 - 5x)/2 = - 5/x
this is a quadratic...
5x - 5x^2 = - 10
5x^2 - 5x - 10 = 0
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
there are your two values for x...y values follow easily...
[Your first solution---(5, -1)---is not correct...the coordinate pair is (- 1, 5)]
Not to make this too dramatic, or to "overstate the case", the word you want is:
HYPERBOLA
[look up hyperbole !]
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First of all, a "hyperbole" is an exaggeration; you mean "hyperbola".
Secondly, I guess when you use a comma (,) is where I would use a period or decimal point (.).
When you substituted y = -2.5x + 2.5 into y = -5/x, you should have had -2.5x + 2.5 = -5/x.
-2.5x^2 + 2.5x = -5
0 = 2.5x^2 - 2.5x - 5
0 = 5x^2 - 5x - 10
0 = x^2 - x - 2
0 = (x - 2)(x + 1)
x - 2 = 0 or x + 1 = 0
x = 2...............x = -1
y = -5/2........... y = 5
(2, -1.5) and (-1, 5) are the intersection points.
Secondly, I guess when you use a comma (,) is where I would use a period or decimal point (.).
When you substituted y = -2.5x + 2.5 into y = -5/x, you should have had -2.5x + 2.5 = -5/x.
-2.5x^2 + 2.5x = -5
0 = 2.5x^2 - 2.5x - 5
0 = 5x^2 - 5x - 10
0 = x^2 - x - 2
0 = (x - 2)(x + 1)
x - 2 = 0 or x + 1 = 0
x = 2...............x = -1
y = -5/2........... y = 5
(2, -1.5) and (-1, 5) are the intersection points.