How do you solve the following system?: $12 x + 33 y = 1, - 4 x + 5 y = - 2$ $12x +33y =1, -4x +5y = -2$

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Answer 1 · 2018-04-16T10:40:01

By arranging the equation

$12 x + 33 y = 1$ $12x + 33y = 1$
$- 12 x + 15 y = - 6$ $-12x + 15y = -6$

(after multiplying 3)

Now add the two equations:

$48 y = - 5$ $48y = -5$

$y = - \frac{5}{48}$ $y=-5/48$

Put this value in the 1st original equation:

$12 x = 1 - (33 \times - \frac{5}{48})$ $12x = 1-(33times-5/48)$

$12 x = 1 + (\frac{165}{48})$ $12x = 1+(165/48)$

$12 x = \frac{48 + 165}{48}$ $12x = (48+165)/48$

$12 x = \frac{213}{48}$ $12x = 213/48$

$12 x = \frac{71}{16}$ $12x = 71/16$

$x = \frac{71}{16 \times 12}$ $x = 71/(16times12)$

$x = \frac{71}{192}$ $x=71/192$

Your solution:

$x = \frac{71}{192}$ $x=71/192$
$y = - \frac{5}{48}$ $y=-5/48$

or

$y = - \frac{20}{192}$ $y=-20/192$

How do you solve the following system?: 12x+33y=1,−4x+5y=−212x +33y =1, -4x +5y = -2