Question

5. Calculate v2, when 5 = 3x’z – y223 + 4x’y+2x-3y – 5
at the point (1, 1, 0).
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Solution:

2x + 5y = 12

We have,

x + y = 3

When y = 0 we have x = 3

When x = 0 we have y = 3

Thus we have the following table giving points on the line x + y = 3

X 0 3

Y 3 0

Now, 2 + 5y = 12

When x = 1, we have

Thus we have the following table giving points on the line 2x + 5y = 12

X 1 -4

Y 2 4

Graph of the equation x + y = 3 and 2x + 5y = 12 is

Graph of the equation

Clearly two lines intersect at a point P (1, 2)

Hence x = 1 and y = 2

 

Question: 2

Solve the following of equation graphically

x - 2y = 5

2x + 3y = 10

Solution:

We have, x - 2y = 5 and 2x + 3y = 10

Now, x - 2y = 5

= x = 5 + 2y

When y = 0 then, x = 5

When y = -2 then, x = 1

Thus, we have the following table giving points on the line x - 2y = 5