How do you solve $2 x - y = 4$ $2x - y = 4$ and $4 x = 2 y + 8$ $4x = 2y + 8$ ?

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How do you solve $2 x - y = 4$ $2x - y = 4$ and $4 x = 2 y + 8$ $4x = 2y + 8$ ?

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Answer 1 · 2015-04-07T09:38:56

Dividing both sides of the second equation by 2, we get

$\frac{4 x}{2} = \frac{2 y}{2} + \frac{8}{2}$ $(4x)/2 = (2y)/2+8/2$

$2 x = y + 4$ $2x = y + 4$

$2 x - y = 4$ $2x-y = 4$

This is the same equation as the first one. It means that there are INFINITE solutions for $x$ $x$ and $y$ $y$ .
Any values of x and y that satisfy the first equation will also satisfy the second one.

For eg. if we choose x = 2, y = 0:

First equation :
$2 x - y = 4$ $2x-y = 4$
$(2 \cdot 2) - 0 = 4$ $(2*2) - 0 = 4$ (which is equal to the Right Hand Side)

Second equation :
$4 x = 2 y + 8$ $4x = 2y + 8$
Left Hand Side is $4 x = 4 \cdot 2 = 8$ $4x = 4*2 = 8$
And the Right Hand Side is $2 y + 8 = (2 \cdot 0) + 8 = 8$ $2y + 8 = (2*0) + 8 = 8$

Both the equations will be satisfied for various such values of $x and y$ $x and y$ like $(3, 2); (4, 6)$ $(3,2); (4,6)$ and so on..

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How do you solve $2 x - y = 4$ $2x - y = 4$ and $4 x = 2 y + 8$ $4x = 2y + 8$ ?

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