How do you solve the following system: $2 x - \frac{y}{2} = 4, 5 x + 2 y = 12$ $2x-y/2=4, 5x+2y=12$ ?

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How do you solve the following system: $2 x - \frac{y}{2} = 4, 5 x + 2 y = 12$ $2x-y/2=4, 5x+2y=12$ ?

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Answer 1 · 2016-02-03T21:25:04

$x = \frac{28}{13}, y = \frac{8}{13}$ $x = 28/13 , y = 8/13$

Explanation:

This is done by the process of elimination but it is not the only method.

Begin by multiplying the first equation by 4 to get the coefficients in front of the $y$ $y$ equal. Multiplying the first equation by 4 gives:

$8 x - 2 y = 16$ $8x -2y =16$

Now we can add both equations together

$(5 x + 2 y) + (8 x - 2 y) = (16) + (12)$ $(5x +2y) + (8x-2y) = (16)+(12)$

So we get:

$13 x = 28 \to x = \frac{28}{13}$ $13x = 28 -> x = 28/13$

Now replace $x$ $x$ with this value in either of the equations and solve for $y$ $y$ . So using the first equation: $2 x - \frac{y}{2} = 4$ $2x-y/2 =4$ .

$\to 2 (\frac{28}{13}) - \frac{y}{2} = 4$ $-> 2(28/13) - y/2 =4$
$\frac{56}{13} - \frac{y}{2} = 4$ $56/13 - y/2 = 4$

Rearrange and you should arrive at:

$y = 2 (\frac{56}{13} - 4) = 2 \frac{56 - 52}{13} = \frac{8}{13}$ $y = 2(56/13 - 4) = 2(56-52)/13=8/13$
$therefore x = 28/13, y = 8/13$

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How do you solve the following system: $2 x - \frac{y}{2} = 4, 5 x + 2 y = 12$ $2x-y/2=4, 5x+2y=12$ ?

1 Answer

Explanation:

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How do you solve the following system: $2 x - \frac{y}{2} = 4, 5 x + 2 y = 12$ $2x-y/2=4, 5x+2y=12$ ?