How do solve the following linear system?: $x - 2 y = - 3, 5 x - 3 y = - 29$ $x-2y = -3 , 5x - 3y = -29$ ?

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How do solve the following linear system?: $x - 2 y = - 3, 5 x - 3 y = - 29$ $x-2y = -3 , 5x - 3y = -29$ ?

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Answer 1 · 2016-04-29T14:18:57

$x = - 7, y = - 2$ $x=-7, y=-2$

Explanation:

$x - 2 y = - 3$ $x-2y=-3$
$5 x - 3 y = - 29$ $5x-3y=-29$

Rearrange the first equation to get $x$ $x$ by itself on the left hand side
$x = 2 y - 3$ $x=2y-3$

Then substitute for $x$ $x$ in the second equation.
$5 (2 y - 3) - 3 y = - 29$ $5(2y-3) -3y = -29$
$10 y - 15 - 3 y = - 29$ $10y-15-3y=-29$
$7 y = - 29 + 15 = - 14$ $7y = -29+15 = -14$
$:.y=-2$

Substituting back into the reformatted first equation gives
$x=2(-2)-3$

$:.x= -7$

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How do solve the following linear system?: $x - 2 y = - 3, 5 x - 3 y = - 29$ $x-2y = -3 , 5x - 3y = -29$ ?

1 Answer

Explanation:

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How do solve the following linear system?: x-2y = -3 , 5x - 3y = -29 x−2y=−3,5x−3y=−29 x-2y = -3 , 5x - 3y = -29 ?

1 Answer

Explanation:

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How do solve the following linear system?: $x - 2 y = - 3, 5 x - 3 y = - 29$ $x-2y = -3 , 5x - 3y = -29$ ?