How do solve the following linear system?: $y = - 3 x + 4, 2 x + 5 y = 1$ $y=-3x+4 , 2x+5y=1$ ?

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Answer 1 · 2016-02-06T13:51:07

$x = \frac{19}{13}$ $x=19/13$
$y = - \frac{5}{13}$ $y=-5/13$

start from the given then rearrange
the equations this way

$3 x + y = 4$ $3x+y=4$ first equation
$2 x + 5 y = 1$ $2x+5y=1$ second equation

$15 x + 5 y = 20$ $15x+5y=20$ after multiplying first equation by 5 -(third equation)

$13 x + 0 = 19$ $13x+0=19$ after subtracting the second from the third equation.

$13 x = 19$ $13x=19$ by simplification

divide by 13 now

$\frac{13 x}{13} = \frac{19}{13}$ $(13x)/13=19/13$ by simplification

$(cancel13x)/cancel13=19/13$ by simplification

$x=19/13$

solve $y$ using first equation by substituting the x value.

$3x+y=4$ first equation

use $x=19/13$

$3(19/13)+y=4$

$y=4-3(19/13)$

$y=(52-57)/13$

$y=-5/13$

have a nice day!!! from the Philippines..

How do solve the following linear system?: y=-3x+4 , 2x+5y=1 y=−3x+4,2x+5y=1 y=-3x+4 , 2x+5y=1 ?