How do you solve the following system of equations?: $13x+13y=16, 3x + 19y=19$ ?

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Answer 1 · 2017-10-22T01:18:59

$x = 57/208, y = 199/208$

$13x + 13y = 16$ ..... $(I)$

$3x + 19y = 19$ ..... $(II)$

Rearranging $(II)$ for x for simplicity,

$3x = 19 - 19y$

$implies$ $x = (19-19y)/3$

Substituting for $x$ in $(I)$

$13( (19-19y)/3) + 13y = 16$

$implies$ $(247 - 247y)/3 + 13y = 16$

$implies$ $(247-247y+39y)/3 = 16$

$implies$ $247-208y = 16*3$

$implies$ $208y = 247 - 48$

$implies$ $y = 199/208$

Substituting for $y$ in either equation (we'll choose $(II)$ ),

$3x + 19(199/208) = 19$

$implies$ $3x = 19 - 3781/208$

$implies$ $x = 1/3*((3952 - 3781)/208) = 1/3 * (171/208) = 171/624 = 57/208$

How do you solve the following system of equations?: 13x+13y=16, 3x + 19y=1913x+13y=16, 3x + 19y=19?