How do you solve the following system: $3 x + 2 y = 0, 7 y - 6 x - 5 = 0$ $3x+2y =0, 7y − 6x − 5 = 0$ ?

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Answer 1 · 2016-04-03T17:12:20

$y = \frac{5}{11}, x = - \frac{10}{33}$ $y = 5/11, x = -10/33$

Rearrange the first equation to give a value of $y$ $y$ with respect to $x$ $x$ .

$3 x + 2 y = 0$ $3x + 2y = 0$
$3 x = - 2 y$ $3x = -2y$
$- \frac{3}{2} x = y$ $-3/2x = y$

Now substitute your new value of $y$ $y$ into the second equation

$7 (- \frac{3}{2} x) - 6 x - 5 = 0$ $7(-3/2x) - 6x - 5 = 0$
$- \frac{21}{2} x - 6 x - 5 = 0$ $-21/2x - 6x - 5 = 0$
#-33x - 5 = 0

Rearrange and solve for $x$ $x$

$- \frac{33}{2} x = 5$ $-33/2x = 5$
$x = - \frac{10}{33}$ $x = -10/33$

And because we know $y$ $y$ with respect to $x$ $x$ , we can solve

$y = - \frac{3}{2} x$ $y = -3/2x$
$y = - \frac{3}{2} \cdot - \frac{10}{33} = \frac{30}{66} = \frac{5}{11}$ $y = -3/2 * -10/33 = 30/66 = 5/11$

Substitute both values into either original equation and check if it's right.

$3 x + 2 y = 0$ $3x + 2y = 0$
$- \frac{30}{33} + \frac{10}{11} = 0$ $-30/33 + 10/11 = 0$

It checks out mathematically, so it must be correct.

How do you solve the following system: 3x+2y=0,7y−6x−5=03x+2y =0, 7y − 6x − 5 = 0 ?