How do solve the following linear system?: $x + 3 y = 9, x - 2 y = 8$ $x + 3y = 9 , x-2y=8$ ?

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Answer 1 · 2016-02-09T08:44:00

Solution is $x = 8 \frac{2}{5} or 8.4$ $x=8 2/5 or 8.4$ and $y = \frac{1}{5} or 0.2$ $y=1/5 or 0.2$

In these pair of linear equations, you can eliminate $x$ $x$ simply by subtracting second equation from first, which gives

$(x + 3 y) - (x - 2 y) = 9 - 8$ $(x+3y)-(x-2y)=9-8$ i.e.

$5 y = 1$ $5y=1$ or $y = \frac{1}{5}$ $y=1/5$ or $y = 0.2$ $y=0.2$

Putting this in second equation $x - 2 (\frac{1}{5}) = 8$ $x-2(1/5)=8$ or

$x = 8 + (\frac{2}{5})$ $x=8+(2/5)$ i.e. $x = 8.4$ $x=8.4$

Hence solution is $x = 8 \frac{2}{5} or 8.4$ $x=8 2/5 or 8.4$ and $y = \frac{1}{5} or 0.2$ $y=1/5 or 0.2$

How do solve the following linear system?: x+3y=9,x−2y=8 x + 3y = 9 , x-2y=8 ?