How do you solve the following system: $7 y - 2 x = 10, 11 y + 3 x = 4$ $7y - 2x = 10, 11y + 3x = 4$ ?

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Answer 1 · 2017-07-17T12:05:42

Arrange your equations, finding $x = - \frac{82}{43}$ $x=-82/43$ and $y = \frac{38}{43}$ $y=38/43$

Enlarge the first equation with 3 and the second with 2:

$21 y - 6 x = 30$ $21y - 6x = 30$
$22 y + 6 x = 8$ $22y + 6x = 8$

Combine these equations:

$21 y + 22 y = 38$ $21y + 22y = 38$

$43 y = 38$ $43y = 38$

$y = \frac{38}{43}$ $y=38/43$

Now you can find x, put this y value into any original equation

$21 \times (\frac{38}{43}) - 6 x = 30$ $21times(38/43) - 6x = 30$

$- 6 x = 30 - (\frac{798}{43})$ $-6x = 30 - (798/43)$

$- 6 x = \frac{492}{43}$ $-6x = 492/43$

$x = - \frac{492}{43 \times 6}$ $x = -492/(43times6)$

$x = - \frac{82}{43}$ $x= -82/43$

How do you solve the following system: 7y - 2x = 10, 11y + 3x = 4 7y−2x=10,11y+3x=47y - 2x = 10, 11y + 3x = 4 ?