Solve the equation using substitution method:
$3 x - 5 y = 11$ and $+ 10 y + 6 x = 7$

(\frac{29}{12}, \frac{3}{4})

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(\frac{3}{4}, \frac{29}{12})

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(\frac{- 29}{12}, \frac{3}{4})

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(\frac{29}{12}, \frac{- 3}{4})

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Solution

The correct option is C $(\frac{29}{12}, \frac{- 3}{4})$
$3 x - 5 y = 11$ ---------- (1)
$+ 10 y + 6 x = 7$ ---------(2)
From equation (1)
$3 x - 5 y = 11$
$\frac{3 x - 11}{5} = y$
Substitute the value of $y$ in equation (2)
$+ 10 (\frac{3 x - 11}{5}) + 6 x = 7$
$+ 6 x - 22 + 6 x = 7$
$12 x = 7 + 22$
$x = \frac{29}{12}$
Substitute the value of $x$ in equation (1)
$3 x - 5 y = 11$
$3 (\frac{29}{12}) - 5 y = 11$
$29 - 20 y = 44$
$29 - 44 = 20 y$
$\frac{- 153}{20} = y$
$y = \frac{- 3}{4}$
therefore solution is $x = \frac{29}{12}$ and $y = \frac{- 3}{4}$

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Simplify:

(1)

(2)

(3)

(4)

(5)

(6)

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