How do you solve the following system: $11 y + 3 x = 4, - 3 y + x = - 3$ $11y + 3x = 4, -3y + x = -3$ ?

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Answer 1 · 2018-03-24T15:03:20

$y = \frac{13}{20}$ $color(blue)(y=13/20)$

$x = - \frac{21}{20}$ $color(blue)(x=-21/20)$

$11y+3x=4 \ \ \ \ \ \ \ \ \ [1]$

$-3y+x=-3 \ \ \ \ [2]$

Multiply equation $[2]$ by 3 and subtract it from equation $[1]$

$11y+3x=4 \ \ \ \ \ \ \ \ \ [1]$

$-9y+3x=-9 \ \ \ \ [2]$

$\ \ \ \ \ \ \ \ \ \ \ \ \20y=13$

Divide by 20:

$color(blue)(y=13/20)$

We now substitute this into one of the original equations, it dosen't matter which one we choose.

In $[2]$

$-3(13/20)+x=-3$

Add $-3(13/20)$ to both sides:

$x=-3+3(13/20)$

$color(blue)(x=-21/20)$

How do you solve the following system: 11y + 3x = 4, -3y + x = -3 11y+3x=4,−3y+x=−311y + 3x = 4, -3y + x = -3 ?