How do you solve the following system?: $2 x + 13 y = 7, 8 x - 7 y = 0$ $2x +13y =7, 8x -7y = 0$

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Answer 1 · 2018-03-06T12:14:05

This can be solved by subtraction.

$x = \frac{49}{118}$ $x=49/118$

$y = \frac{28}{59}$ $y=28/59$

Multiplying both sides by 4,
$4 (2 x + 13 y) = 4 (7)$ $4(2x + 13y) = 4(7)$

$8 x + 52 y = 28$ $8x + 52y = 28$ --- (1)

$8 x - 7 y = 0$ $8x - 7y = 0$ --- (2)

Subtract the two equations:
(1) - (2): $8 x + 52 y - 8 x + 7 y = 28 - 0$ $8x +52y - 8x + 7y = 28 - 0$

$59 y = 28$ $59y = 28$
$y = \frac{28}{59}$ $y = 28/59$

$x = \frac{7 y}{8}$ $x = (7y)/8$
$x = 7 (\frac{28}{59}) \cdot (\frac{1}{8}) = \frac{49}{118}$ $x = 7(28/59)*(1/8) = 49/118$

How do you solve the following system?: 2x+13y=7,8x−7y=02x +13y =7, 8x -7y = 0