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Manipulation of a linear equation

Which of the ...

Question

Which of the following is not a solution of the pair of equations $3 x - 2 y = 4$ and $6 x - 4 y = 8$ ?

x = 2, y = 1

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x = 4, y = 4

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x = 6, y = 7

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x = 5, y = 3

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Solution

The correct option is D $x = 5, y = 3$
As the value of $x (= 5)$ and $y (= 3)$ does not satisfies any of the given equation, hence $(5, 3)$ is not a solution of the pair of equations $3 x - 2 y = 4$ and $6 x - 4 y = 8$ .

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Similar questions

x = 2, y = 1

x = 4, y = 4

3 x - 2 y = 4

6 x - 4 y = 8

Find out which of the following equations have x=1, y=2 as solution.
x + 2y = 5
2y - 3x + 3 = 0
3x + 4y - 8 = 0 [4 MARKS]

\frac{1}{2 x} + \frac{1}{3 y} = 2; \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2; \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1

\frac{4}{x} + 3 y = 14; \frac{3}{x} - 4 y = 23

\frac{5}{x - 1} + \frac{1}{y - 2} = 2; \frac{6}{x - 1} - \frac{3}{y - 2} = 1

\frac{7 x - 2 y}{x y} = 5; \frac{8 x + 7 y}{x y} = 15

6 x + 3 y = 6 x y; 2 x + 4 y = 5 x y

\frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = - 2

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}; \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}

\frac{1}{2 x} - \frac{1}{y} = - 1

\frac{1}{x} + \frac{1}{2 y} = 8

x \neq 0 a n d y \neq 0)

\frac{1}{2 x} - \frac{1}{y} = - 1

\frac{1}{x} + \frac{1}{2 y} = 8

x \neq 0, y \neq 0)

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