Question

4x+2/3y-1=0 and 6x-y+2

Answer 1

$\frac{4x + 2}{3y - 1} = 0 \\ \\ 4x + 2 = 0(3y - 1) = 0$

If 4x +2 = 0, then 3y - 1 would contain all positive numbers.

$4x + 2 = 0 \\ \\ 4x = - 2 \\ \\ x = \frac{- 2}{4} = \frac{- 1}{2}$

Least value of 3y - 1 is 1.

$3y - 1 = 1 \\ \\ 3y = 1 + 1 = 2 \\ \\ y = \frac{2}{3} \\ \\ \\ 3y - 1 = 2 \\ \\ 3y = 2 + 1 = 3 \\ \\ y = \frac{3}{3} = 1 \\ \\ \\ 3y - 1 = 3 \\ \\ 3y = 3 + 1 = 4 \\ \\ y = \frac{4}{3}$

∴ The value of y is one among 2/3, 1, 4/3, 5/3, 2, 7/3,......

The least value of y is 2/3.

Let's find the least value of 6x - y + 2.

$6x - y + 2 \\ \\ = 6(\frac{-1}{2}) - \frac{2}{3} + 2 \\ \\ = - 3 - \frac{2}{3} + 2 \\ \\ = - 1 - \frac{2}{3} = - \frac{5}{3}$

$6x - y + 2 \\ \\ = 6(\frac{-1}{2}) - 1 + 2 \\ \\ = - 3 - 1 + 2 \\ \\ = - 2$

$6x - y + 2 \\ \\ = 6(\frac{-1}{2}) - \frac{4}{3} + 2 \\ \\ = - 3 - \frac{4}{3} + 2 \\ \\ = - 1 - \frac{4}{3} = - \frac{7}{3}$

∴ The value of 6x - y + 2 is one among -(5/3), -2, -(7/3), -(8/3), -3, -(10/3),.....

The least value of 6x - y + 2 is -(5\3).

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