How do you solve the following linear system $2 x + 3 y = 3, 3 x + y = 3$ $2x+3y= 3, 3x+y=3$ ?

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Answer 1 · 2016-02-10T00:35:04

The easiest way in this situation would be substitution, and so I'll solve it that way for you.

To do substitution, you must isolate a variable.

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

$y = 3 - 3 x$ $y = 3 - 3x$

$2 x + 3 (3 - 3 x) = 3$ $2x + 3(3 - 3x) = 3$

$2 x + 9 - 9 x = 3$ $2x + 9 - 9x = 3$

$- 7 x = - 6$ $-7x = -6$

$x = \frac{6}{7}$ $x = 6/7$

Plugging in $\frac{6}{7}$ $6/7$ for x.

$3 (\frac{6}{7}) + y = 3$ $3(6/7) + y = 3$

$y = 3 - \frac{18}{7}$ $y = 3 - 18/7$

$y = \frac{3}{7}$ $y = 3/7$

The solution set is ${\frac{6}{7}, \frac{3}{7}}$ ${6/7, 3/7}$

Hopefully this helps!

How do you solve the following linear system 2x+3y= 3, 3x+y=3 2x+3y=3,3x+y=32x+3y= 3, 3x+y=3 ?