How do you solve the following system: $8 x - 3 y = 3, 5 x + 8 y = - 29$ $8x-3y=3, 5x+8y= -29$ ?

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Answer 1 · 2017-07-24T07:56:58

Arrange equations, you will get $x = - \frac{63}{79}$ $x=-63/79$ and $y = - \frac{247}{79}$ $y=-247/79$

Enlarge the first equation by 8 and the second equation by 3.

$64 x - 24 y = 24$ $64x - 24y = 24$
$15 x + 24 y = - 87$ $15x + 24y = -87$

Now you have (after adding these equations):
$79 x = - 63$ $79x = -63$

Now you can have $x = - \frac{63}{79}$ $x=-63/79$

Find y now after putting x value in the first original equation:

$8 \times (- \frac{63}{79}) - 3 y = 3$ $8times(-63/79) -3y = 3$

$- 3 y = 3 + (\frac{504}{79})$ $-3y = 3 + (504/79)$

$- 3 y = \frac{741}{79}$ $-3y = 741/79$

$y = - \frac{247}{79}$ $y= -247/79$

Now you have y

$y = - \frac{247}{79}$ $y=-247/79$

How do you solve the following system: 8x-3y=3, 5x+8y= -29 8x−3y=3,5x+8y=−298x-3y=3, 5x+8y= -29 ?