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Value of an Expression

Which one of ...

Question

Which one of the following is not a linear equation?

2x – 5 = 8

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2x + 2y = 8

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23x + z = p + q

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p = 23 (x + 1) (x - 1)

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Solution

The correct option is D
p = 23 (x + 1) (x - 1)

(x + 1) (x - 1) = $x^{2}$ – 1 (Not linear)

Suggest Corrections

Similar questions

\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

\frac{x - 5}{1} = \frac{2 y + 6}{- 2} = \frac{z - 3}{1} and \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

The equation of the set of points P which is equidistant from the points (0, 2, 3) and (2, –2, 1) is:

\frac{x - 1}{2} = \frac{y - 1}{1} = \frac{z - 0}{4}

x + 2 y - z = 3

P

x + 2 y + 3 z + 1 = 0 = x - y - z - 6,

- 2 x + y + z + 8 = 0.

P ?

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