How do you solve the following system?: $2x +3y =5 , 8x +5y = -4$

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Answer 1 · 2018-04-21T23:45:33

$X = 33/16 and Y = 48/14$

Given:
$2x+3y = 5 and 8x + 5y = -4$
We need to make either one of the terms given, have similar coefficients to cancel out.

$(2x + 3y = 5)$ x $8$
$color(red)(16x + 24y = 40)$

$(8x+ 5y = -4)$ x $2$
$color(red)(16x + 10y = -8)$

Subtracting one from the other, we get:
$16x + 24 y = 40$
$-(16x+10y = -8)$

We get $14y = 48; y =48/14$

Substituting y = $48/14$ in any one of the equations.
$2x + cancel(3)(14/cancel(48)) = 5$
$2x + 7/8 = 5$
$2x = 5-7/8$
$2x = (40-7)/8$
$:. x = 33/16$

How do you solve the following system?: 2x +3y =5 , 8x +5y = -42x +3y =5 , 8x +5y = -4