Question

find the four different solutions of the equation 3x+7y=5​

Answer 1

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1. x = 3 , y = 2

2. x = 4 , y = -1

3. x = 5/3 , y = 0

4. x = 19/3 , y = -2

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${\huge{\underline{\sf {\pink{Explanation :}}}}}$

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Given :

$\sf3x + 7y = 5$

$\sf3x = 5 - 7y$

$\sf \: x = \frac{5 \: - \: 7y}{3}$

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Putting any value of y we will get the value of x that would satisfy the given equation.

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Let, y = 2

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$\sf \: x \: = \frac{5 \: - \: 7(2)}{3}$

$\sf x \: = \frac{5 - 14}{3}$

$\sf \: x = \frac{9}{3}$

$\sf \: x = 3$

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Let, y = -1

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$\sf \: x \: = \frac{5 \: - \: 7(-1)}{3}$

$\sf \: x \: = \frac{(5 \: + 7)}{3}$

$\sf \: x \: = \frac{12)}{3}$

$\sf \: x = 4$

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Let, y = 0

$\sf \: x \: = \frac{5 \: - \: 7(0)}{3}$

$\sf \: x \: = \frac{5}{3}$

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Let, y = -2

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$\sf \: x \: = \frac{5 \: - \: 7(-2)}{3}$

$\sf \: x \: = \frac{(5 \: + 14)}{3}$

$\sf \: x \: = \frac{19}{3}$

Answer 2

Step-by-step explanation:

3x+7y=5

• hence ,R.HS is not equal to LHs