How do you solve the system $2 x + 3 y = 2$ $2x + 3y = 2$ , $x + 6 y = 4$ $x + 6y = 4$ ?

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How do you solve the system $2 x + 3 y = 2$ $2x + 3y = 2$ , $x + 6 y = 4$ $x + 6y = 4$ ?

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Answer 1 · 2016-01-11T21:04:59

$y = \frac{2}{3}$ $y=2/3$ and $x = 0$ $x=0$

Explanation:

Multiply both sides of the second equation by 2 and subtract this new equation from the first equation. One gets the equation $- 9 y = - 6$ $-9y = -6$ , or, equivalently, $y = \frac{6}{9} = \frac{2}{3}$ $y = 6/9 = 2/3$ Substituting this value for y in equation 2 gives: $x + 6 (\frac{2}{3}) = 4$ $x + 6(2/3) = 4$ , or $x - 4 = 4$ $x-4=4$ . So, x = 0.

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How do you solve the system $2 x + 3 y = 2$ $2x + 3y = 2$ , $x + 6 y = 4$ $x + 6y = 4$ ?

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How do you solve the system $2 x + 3 y = 2$ $2x + 3y = 2$ , $x + 6 y = 4$ $x + 6y = 4$ ?