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28 Answers

in this case, the LHS of both equations differ by a factor of 2 and the right hand side is not, so there is no solution. Geographically it means that the lines represented by the eqns are parallel and has no intersection
by Level 3 User (2.1k points)
5x+4y=-30

3x-9y=-18
by Level 1 User (140 points)
2.1x - 0.9y =15,-1.4x + 0.6y=10
by
4x - 4y = 2

7x + 2y = 17
by
-x-5y-5z=2

4x-5y+4z=19

x+5y-z=-20
by
solve linear equation by gauss elimination methhod
how do u solve Xsquared= 5y?
by
2(4x+y)=22 8X+2Y=22 8X+2Y-8X+2Y=22-11 Y=11
by
X=0 and y=6
by
8x+2y=12 when x=0 2y=12 y=12/2 y=6 coordinates for x & y are (0,6) when y=0 8x=12 x=12/8 x=3/2 x=1.1/2 coordinates for x & y are (1.1/2,0) 4x+y=11 when x=0 y=11 coordinates for x & y are (0,11) when y=0 4x=11 x=11/4 x=2.3/4 coordinates for x & y are (2.3/4,0)
by
8x + 2y = 12.........(1) 4x + y = 11.........(2) Multiplying eq. 1 by 2 and eq.2 by 1, you get 8x + 2y = 12 8x + 2y = 22 Therefore 8x = 12 - 2y, and 8x = 22 - 2y There will never be a SOLUTION here
by
a^2+c^2 a=2 c=4
by
no ANswer
by
A coffee manufacturer is interested in blending three different types of coffee beans into a final coffee blend. The three component beans cost hte manufacturer Rs. 1.2, Rs. 1.6 and Rs. 1.4 per kilogram, respectively. the manufacturer wants to blend a batch of 40,000 kg of coffee and has a coffee-purchasing budget of Rs. 57, 600. In blending the cofee, one restriction is that the amount used of component 2 should be twice that of comonent 1, for a better flavour. Determine which combination of the three components will yield a final blend, using Gasussian elimination method.
by
4x + 3y =4

2x - 5y =15
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8x+2y=12
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1/2y^2 - 5/2y - 12 = 1/2y(y-?)(y+?)
by
3x+2y=6

x-3y=13
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x= ~2y

x+4=2
by
-4x-2y=-12 4x+8y=-24
by
4xz+8y=20 -4x+2y=-30
by
4x+8y=20 -4x+2y+-30
by Level 1 User (140 points)
lx+my=1 ;(l+m)x2+a=0
by
y=2x-2 y=x+1
by
8x+2y=12 4x+y=11

(4x)-11y
by
X=-10 and Y=51
by
y =4       name is simone
by
5 X-Y=9

Y=-2X=9
by
Y=10.667 and X=undifined
by

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