Question

the area enclosed by
$3 |x| + 2 |y| \leqslant 6$
Is ​

Answer 1

Answer:

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Answer 2

Answer:

Let us learn the rule first.

Two equations are -

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

Then the solution be presented as

x/(b₁c₂ - b₂c₁) = y/(a₂c₁ - a₁c₂) = 1/(a₁b₂ - a₂b₁)

Given equations are

2x + 3y - 26 = 0 ...(i)

3x - 2y + 0 = 0 ...(ii)

From (i) and (ii), using Cross-multiplucation method, we get

x/(0 - 52) = y/(- 78 - 0) = 1/(- 4 - 9)

or, x/(- 52) = y/(- 78) = 1/(- 13)

or, x/4 = y/6 = 1

Then, x = 4 and y =6

Step-by-step explanation:

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