Question

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Find three solutions each , of the following linear equations :-

$1) \: 7x + 5y = 46$

$2) \: 5x + 7y = 50$

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Answer 1

1.

Given equation

    7x + 5y = 46

➙ $\frac{7}{46}x+\frac{5}{46}y=1$

➙ $\dfrac{x}{\frac{46}{7}}+\dfrac{y}{\frac{46}{5}}=1$

  Using graphical representation of straight lines in two dimensions, we can say that the points $(\frac{46}{7},0)$ and $(0,\frac{46}{5})$ lie on the line 7x + 5y = 46.

Therefore, two solutions can be

$x=\frac{46}{7}\:,\:y=0\:and\:x=0\:,y=\frac{46}{5}$

Putting x = 1 in the given equation, we can find that

    (7 * 1) + 5y = 46

    ➙ 7 + 5y = 46

    ➙ 5y = 46 - 7

    ➙ y = $\frac{39}{5}$

So, another solution be

$x=1\:,\:y=\frac{39}{5}$

Therefore, the three solutions are

$\boxed{\tiny{\bold{\boxed{x=\frac{46}{7}\:,\:y=0}\:;\:\boxed{x=0\:,\:y=\frac{46}{5}}\:and\:\boxed{x=1\:,\:y=\frac{39}{5}}}}}$

2.

Given equation

    5x + 7y = 50

➙ $\frac{5}{50}x+\frac{7}{50}y=1$

➙ $\dfrac{x}{10}+\dfrac{y}{\frac{50}{7}}=1$

  Using graphical representation of straight lines in two dimensions, we can say that the points (10, 0) and $(0,\frac{50}{7})$ lie on the line 5x + 7y = 50.

Therefore, two solutions can be

$x=10\:,\:y=0\:and\:x=0\:,y=\frac{50}{7}$

Putting x = 1 in the given equation, we can find that

    (5 * 1) + 7y = 50

    ➙ 5 + 7y = 50

    ➙ 7y = 50 - 5

    ➙ y = $\frac{45}{7}$

So, another solution be

$x=1\:,\:y=\frac{45}{7}$

Therefore, the three solutions are

$\boxed{\tiny{\bold{\boxed{x=10\:,\:y=0}\:;\:\boxed{x=0\:,y=\frac{50}{7}}\:and\:\boxed{x=1\:,\:y=\frac{45}{7}}}}}$