How do you solve the system of equations $2 x - 5 y = 3$ $2x-5y=3$ and $x - 3 y = 1$ $x-3y=1$ ?

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How do you solve the system of equations $2 x - 5 y = 3$ $2x-5y=3$ and $x - 3 y = 1$ $x-3y=1$ ?

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Answer 1 · 2016-09-30T21:56:49

Try to get rid of one of the unknown...

Explanation:

$2 x - 5 y = 3$ $2x-5y=3$
$x - 3 y = 1$ $x-3y=1$ multiply this one by - 2 and add to first one so $x$ $x$ term will disappear and we will get an equation with only one unknown parameter which is $y$ $y$

$2 x - 5 y = 3$ $2x-5y=3$
$- 2 x + 6 y = - 2$ $-2x+6y=-2$
$0 + y = 1 \Rightarrow y = 1$ $0+y=1 => y=1$
Put 1 instead of $y$ $y$ in one of the equations to obtain $x$ $x$
$x - 3 \cdot (1) = 1$ $x-3*(1)=1$
$x - 3 = 1 \Rightarrow x = 4$ $x-3=1 =>x=4$

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How do you solve the system of equations $2 x - 5 y = 3$ $2x-5y=3$ and $x - 3 y = 1$ $x-3y=1$ ?

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Explanation:

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How do you solve the system of equations $2 x - 5 y = 3$ $2x-5y=3$ and $x - 3 y = 1$ $x-3y=1$ ?