Problem 1.
-4x+8y= 1
3x-6y= 1
Problem 2.
2x+5y= 15
-6x-15y= -45
2006-07-18 04:55:02 · 10 answers · asked by Brandon ツ 3 in Science & Mathematics ➔ Mathematics
Look at my record, I do will select an answer asap.
2006-07-18 04:55:57 · update #1
-4x + 8y = 1
3x - 6y = 1
-4x + 8y = 1
8y = 4x + 1
y = (1/2)x + (1/8)
3x - 6y = 1
-6y = -3x + 1
y = (1/2)x - (1/6)
Same slope, different y-intercepts
ANS : No Solution
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2x + 5y = 15
-6x - 15y = -45
2x + 5y = 15
5y = -2x + 15
y = (-2/5)x + 3
-6x - 15y = -45
-15y = 6x - 45
y = (-6/15)x + 3
y = (-2/5)x + 3
Since they have the same slope and same y-intercept
ANS : Infinite Solutions
2006-07-18 07:13:53 · answer #1 · answered by Sherman81 6 · 4⤊ 0⤋
There are no solutions to the first problem:
[-4x+8y=1] * 3 => -12x + 24y =3
[3x-6y=1] * 4 => 12x - 18y = 3
adding the equations yields: 0 + 6y = 0 so y is 0 but there is no number that when multiplied by -4 and 3 yields 1
[2x + 5y=15] * 3 => 6x + 15y = 45
-6x - 15y =-45 =>-6x - 15y = -45
adding the equations yields 0 = 0, but if x is 0, y = 3
2006-07-18 12:12:22 · answer #2 · answered by Wytecyde 1 · 0⤊ 0⤋
For problem 2: they are the equation really. multiply the top by -3 and you have the same line. so the solution is the line of the equation.
problem 1:
add the two equations: -x + 2y = 2 ==> x=2(y-1)
plug into bot. eqn: 3(2(y-1))-6y=1 ==> 1 = 1
thus there's no intersection of the two lines, they are parallel lines
2006-07-18 12:07:53 · answer #3 · answered by cw 3 · 0⤊ 0⤋
Note that a linear equation represents a straight line in the x-y space.
First one:
they are parallel lines: so no solution
Second one:
they are coincident: so infinite solutions (all points lying on the line 2x+5y=15)
2006-07-18 12:03:27 · answer #4 · answered by robo 1 · 0⤊ 0⤋
Empty set for the first problem. They work out to having the same slope but different y-intercepts.
y=1/2x+1/8
y=1/2x-1/6
Second system works out to infinity. They work out to being the same line with formula y=2/5x+3
2006-07-18 12:09:39 · answer #5 · answered by carpetao 3 · 0⤊ 0⤋
Both sets of equations are underdetermined.
In problem 2 eqaution 2 equals equation one multiplied by -3
In problem one equation two equals equation one multiplied
by -3/4.
You need two linearly independent equations to solve for x and Y.
You do not have them in these examples
2006-07-18 12:07:33 · answer #6 · answered by gilbraltor 1 · 0⤊ 0⤋
Skywalker, simultaneous as well as quadratic equations are lots of fun. Pay closer attention in class. You'll enjoy them.
2006-07-18 12:32:14 · answer #7 · answered by Njihem 2 · 0⤊ 0⤋
They are parrallel lines. They meet at infinity. So x= y = infinity is one mathematical solution.
2006-07-18 12:37:07 · answer #8 · answered by Dr M 5 · 0⤊ 0⤋
try r
2006-07-18 12:02:44 · answer #9 · answered by sahi 2 · 0⤊ 0⤋
Please do your OWN homework!
2006-07-18 12:29:10 · answer #10 · answered by ? 3 · 0⤊ 0⤋