How do you solve the linear system by substitution 3x + y =2 and -x -3y=6?

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Answer 1 · 2018-04-28T19:45:09

Like so:

$3 x + y = 2$ $3x+y=2$

$- x - 3 y = 6 \Rightarrow - 6 - 3 y = x$ $-x-3y=6 => -6-3y=x$

Substitute this equation into the other system wherever you see $x$ $x$ .

$3 (- 6 - 3 y) + y = 6$ $3(-6-3y)+y=6$

Distribute $3$ $3$ onto $(- 6 - 3 y)$ $(-6-3y)$

$- 18 - 9 y + y = 2$ $-18-9y+y=2$

Combine alike terms

$- 8 y = 20$ $-8y= 20$

Divide to solve for $y$ $y$

$\frac{- 8 y}{- 8} = \frac{20}{- 8} \Rightarrow y = - \frac{20}{8}$ $(-8y)/-8=20/(-8) => y= -20/8$

Then solve for $x$ $x$ using the value of $y$ $y$ .

$- x - \frac{60}{8} = 6$ $-x-60/8 = 6$

$x = - \frac{108}{8}$ $x=-108/8$

$y = - \frac{20}{8}$ $y = -20/8$

Hope that helps!