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

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

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Answer 1 · 2017-10-22T01:18:59

$x = \frac{57}{208}, y = \frac{199}{208}$ $x = 57/208, y = 199/208$

Explanation:

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

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

Rearranging $(I I)$ $(II)$ for x for simplicity,

$3 x = 19 - 19 y$ $3x = 19 - 19y$

$\Rightarrow$ $implies$ $x = \frac{19 - 19 y}{3}$ $x = (19-19y)/3$

Substituting for $x$ $x$ in $(I)$ $(I)$

$13 (\frac{19 - 19 y}{3}) + 13 y = 16$ $13( (19-19y)/3) + 13y = 16$

$\Rightarrow$ $implies$ $\frac{247 - 247 y}{3} + 13 y = 16$ $(247 - 247y)/3 + 13y = 16$

$\Rightarrow$ $implies$ $\frac{247 - 247 y + 39 y}{3} = 16$ $(247-247y+39y)/3 = 16$

$\Rightarrow$ $implies$ $247 - 208 y = 16 \cdot 3$ $247-208y = 16*3$

$\Rightarrow$ $implies$ $208 y = 247 - 48$ $208y = 247 - 48$

$\Rightarrow$ $implies$ $y = \frac{199}{208}$ $y = 199/208$

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

$3 x + 19 (\frac{199}{208}) = 19$ $3x + 19(199/208) = 19$

$\Rightarrow$ $implies$ $3 x = 19 - \frac{3781}{208}$ $3x = 19 - 3781/208$

$\Rightarrow$ $implies$ $x = \frac{1}{3} \cdot (\frac{3952 - 3781}{208}) = \frac{1}{3} \cdot (\frac{171}{208}) = \frac{171}{624} = \frac{57}{208}$ $x = 1/3*((3952 - 3781)/208) = 1/3 * (171/208) = 171/624 = 57/208$

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

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Explanation:

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

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Explanation:

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