How do solve the following linear system?: $3 x - 2 y = 2, - 4 x - 14 y = 28$ $3x-2y=2 , -4x-14y=28$ ?

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How do solve the following linear system?: $3 x - 2 y = 2, - 4 x - 14 y = 28$ $3x-2y=2 , -4x-14y=28$ ?

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Answer 1 · 2016-02-02T14:01:31

Solve by substitution and elimination:

$3 x - 2 y = 2$ $3x-2y=2$

$- 4 x - 14 y = 28$ $-4x-14y=28$

We can eliminate $- 14 y$ $-14y$ from the second equation by $- 2 y$ $-2y$ from the first equation if we multiply it with $- 7$ $-7$ to get $14 y$ $14y$

$\to - 7 (3 x - 2 y = 2)$ $rarr-7(3x-2y=2)$

Use distributive property:

$\to - 21 x + 14 y = - 14$ $rarr-21x+14y=-14$

Add both equations:

$\to (- 4 x - 14 y = 28) + (- 21 x + 14 y = - 14)$ $rarr(-4x-14y=28)+(-21x+14y=-14)$

$\to - 25 x = 14$ $rarr-25x=14$

$\to x = - \frac{14}{25}$ $rarrx=-14/25$

Substitute the value of $x$ $x$ to the first equation:

$\to 3 (- \frac{14}{25}) - 2 y = 2$ $rarr3(-14/25)-2y=2$

$\to - \frac{42}{25} - 2 y = 2$ $rarr-42/25-2y=2$

$\to - 2 y = 2 + \frac{42}{25}$ $rarr-2y=2+42/25$

$\to - 2 y = \frac{92}{25}$ $rarr-2y=92/25$

$\to y = \frac{92}{25} \div (- \frac{2}{1})$ $rarry=92/25-:(-2/1)$

$\to y = \frac{92}{25} \cdot (- \frac{1}{2})$ $rarry=92/25*(-1/2)$

$\to y = - \frac{46}{25}$ $rarry=-46/25$

So,
$(x, y) = (- \frac{14}{25}, - \frac{46}{25})$ $(x,y)=(-14/25,-46/25)$ :)

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How do solve the following linear system?: $3 x - 2 y = 2, - 4 x - 14 y = 28$ $3x-2y=2 , -4x-14y=28$ ?

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How do solve the following linear system?: 3x−2y=2,−4x−14y=28 3x-2y=2 , -4x-14y=28 ?

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How do solve the following linear system?: $3 x - 2 y = 2, - 4 x - 14 y = 28$ $3x-2y=2 , -4x-14y=28$ ?