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

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Answer 1 · 2016-09-03T10:28:19

$x = \frac{49}{22}$ $x=49/22$

$y = \frac{28}{11}$ $y=28/11$

$2 x + y = 7$ $2x+y=7$ or $8 x + 4 y = 28$ $8x+4y=28$

$8 x - 7 y = 0$ $8x-7y=0$

or

$8 x = 7 y$ $8x=7y$

Plugging the value $8 x = 7 y$ $8x=7y$ in the equation

$8 x + 4 y = 28$ $8x+4y=28$

We get

$7 y + 4 y = 28$ $7y+4y=28$

or

$11 y = 28$ $11y=28$

or

$y = \frac{28}{11}$ $y=28/11$ ------------------------------Ans $1$ $1$

Plugging the value $y = \frac{28}{11}$ $y=28/11$ in the equation $2 x + y = 7$ $2x+y=7$

We get

$2 x + \frac{28}{11} = 7$ $2x+28/11=7$

or

$2 x = 7 - \frac{28}{11}$ $2x=7-28/11$

or

$x = \frac{49}{11} \times \frac{1}{2}$ $x=49/11times1/2$

or

$x = \frac{49}{22}$ $x=49/22$ ----------------------------------Ans $2$ $2$

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