Question

consider the following linear equation for 4x+ 3y - 4z=3x-y + Z = 12x-y= 1 the value of x is​

Answer 1

Answer:

Correct option is

B

(120,0)

Given constraints, 4x+3y≤480 and 2x+3y≤360

first, draw the graph for equations 4x+3y=480 and 2x+3y=360

for 4x+3y=480

substitute y=0 we get, 4x=480⟹x=120

substitute x=0 we get, 3y=480⟹y=160

therefore, 4x+3y=480 line passes through (120,0) and (0,160) as shown in fig.

Hence, 4x+3y≤480 includes the region below the line.

for 2x+3y=360

substitute y=0 we get, 2x=360⟹x=180

substitute x=0 we get, 3y=360⟹y=120

therefore, 4x+3y=480 line passes through (180,0) and (0,120) as shown in fig.

Hence, 2x+3y≤360 includes the region below the line.

the shaded region as shown in figure is intersection region. From the figure, (0,120) and (120,0) are feasible corner points.

Answer 2

Final answer: x=1  

Given that: We are given 4x+3y-4z = 3x-y+Z = 12x-y= 1

To find: We have to find x.

Explanation:

• 4x+ 3y- 4z = 3x-y+Z = 12x-y = 1 which means,

        4x + 3y- 4z = 1

        3x-y+z = 1

        12x – y = 1

• 12x-y = 1 from this y = 12x-1

• Substitute y = 12x-1 in 4x + 3y- 4z = 1 and 3x-y+z = 1
• 4x + 3y- 4z = 1 become 4x + 3(12x - 1)-4z = 1

        4x + 36x - 3 - 4z = 1

        40x - 4z - 3 = 1

        40x - 4z = 4

• 3x - y + z = 1 become 3x - (12x - 1) + z = 1

        3x - 12x + 1 +z = 1

        -9x + z + 1 = 1

        z -9x = 0

• Multiplying z-9x=0 with 4, we get 4z - 36x = 0

• Add 4z - 36x = 0  with 40x - 4z = 4

       (40x – 4z = 4) + (-36x + 4z=0)

       = [(40x -36x) + (-4z +4z) = (0 + 4)]

       = 4x = 4

       x = 4/4 = 1

• Put x=1 in 12x – y =1

       (12 * 1) – y =1

       12 – y = 1

       y = 12 - 1 = 11

• Put x=1 and y=11 in 3x - y + z = 1

       (3 * 1) – 11 + z = 1

       3 - 11 + z = 1

       -8 + z = 1

       z = 8 + 1 = 9

• x=1, y=11, z=9

To know more about the concept please go through the links

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