How do you solve the following linear system $3 x - 7 y = 1, x + 3 y = 1 ?$ $3x-7y=1, x+3y=1?$ ?

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Answer 1 · 2018-06-23T15:32:41

Multiply the second equation by $3$ $3$ (everything) and then subtract the second equation from the first

So if you multiply the 2nd equation by $3$ $3$ , you have

$3 x + 9 y = 3$ $3x+9y=3$

The first one is

$3 x - 7 y = 1$ $3x-7y=1$

Now, after subtracting them. we have:

$16 y = 2$ $16y=2$

so

$y = \frac{1}{8}$ $y=1/8$

Now simply put it into any equation from the initial part (let's say the second). We have

$x + 3 \cdot \frac{1}{8} = 1$ $x+3*1/8=1$

So

$x = \frac{5}{8}$ $x=5/8$

$y = \frac{1}{8}$ $y=1/8$

is the only solution.

How do you solve the following linear system 3x−7y=1,x+3y=1?3x-7y=1, x+3y=1? ?