How do you solve the system 3x + y = 5 and -2x + 3y =4?

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Answer 1 · 2017-06-06T08:49:31

$x = 1$ $x = 1$
$y = 2$ $y = 2$

First you would multiply each equation by the coefficient of x

$- 2 (3 x + y = 5)$ $-2(3x+y = 5)$
$3 (- 2 x + 3 y = 4)$ $3(-2x + 3y = 4)$

Which is simplified to

$- 6 x - 2 y = - 10$ $-6x - 2y = -10$
$- 6 x + 9 y = 12$ $-6x + 9y = 12$

Then you would minus both equations together cancelling $x$ $x$ to get the value of $y$ $y$

$- 11 y = - 22$ $-11y = -22$

$y = \frac{22}{11}$ $y= 22/11$

$y = 2$ $y = 2$

Put 2 in replace of $y$ $y$ into one of the original equation to find x

$3 x + 2 = 5$ $3x + 2 = 5$

$3 x = 5 - 2$ $3x = 5 - 2$

$x = \frac{5 - 2}{3}$ $x = (5 - 2)/3$

$x = 1$ $x = 1$