he system of linear equations below has a unique solution for all but one value of a:
7 x−14 y=−14
32 x+a y=−64
What is this exceptional value for a?
2007-03-02 19:14:46 · 6 answers · asked by C 2 in Science & Mathematics ➔ Mathematics
it is when the determinant of the system = 0
thus 7a + 32*14 = 0 so for a = -(2^8) there is not a unique solution
2007-03-02 19:25:59 · answer #1 · answered by gjmb1960 7 · 0⤊ 0⤋
a = -64 makes the second equation read 32x - 64y = -64. Reduced, this reads x - 2y = -2.
7x - 14y = -14. Reduced this also reads x-2y = -2.
When you solve for x, you get x = -2 + 2y
So, if you draw up a small x,y grid:
x y
0 1
2 2
4 3
6 4
plugging any of these values into the equation where a = -64 will result in a valid equation, so there is no "unique" solution
2007-03-03 03:48:57 · answer #2 · answered by n_arent 3 · 0⤊ 0⤋
a= -64 means that the second equation is a multiple of the first. So there are no unique solutions
2007-03-03 03:22:09 · answer #3 · answered by gumtrees 3 · 0⤊ 0⤋
if this has a unique sol.. then a1/a2 will not be equal to b1/b2
ie
7/32is not equal to-14/a
1/4 isnt equal to -14/a
.
. . a is not equal to-64.....
well thats my ans...... hope its right
2007-03-03 03:36:11 · answer #4 · answered by cutie frooty 2 · 0⤊ 0⤋
7x - 14y = -14 (we multiply by 32)
32x - ay = -64 (we multiply by 7)
So, :
224x - 448y = -448
224x + 7ay = -448
_______________ - (we substract it/elimination)
0 - 448y + 7ay = 0
-448y = -7ay
-448y/y = -7a
so:
-448 = -7a
a = 64
2007-03-03 04:37:54 · answer #5 · answered by Chreonne 2 · 0⤊ 0⤋
7x-14y=-14
x-2y=-2
x=2y-2
32x+ay=-64
32(2y-2)+ay=-64
64y-64+ay=-64
64y+ay=0
y(64+a)=0
y=0 or a=-64
When y=0, x= -2
32x+ay=-64
a=0.
2007-03-03 03:33:01 · answer #6 · answered by Xanana 3 · 0⤊ 0⤋