How do you solve the following system: $5 x + 8 y = - 2, 4 x + 7 y = 6 x$ $5x + 8y = -2, 4x + 7y = 6x$ ?

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Answer 1 · 2017-08-05T23:13:56

$x = - \frac{14}{39} and y = - \frac{4}{39}$ $x = - frac(14)(39) and y = - frac(4)(39)$

We have: $5 x + 8 y = - 2 and 4 x + 7 y = 6 x$ $5 x + 8 y = - 2 and 4 x + 7 y = 6 x$

Let's solve both equations for $y$ $y$ :

$\Rightarrow 8 y = - 5 x - 2 and 7 y = 6 x - 4 x$ $Rightarrow 8 y = - 5 x - 2 and 7 y = 6 x - 4 x$

$\Rightarrow y = \frac{- 5 x - 2}{8} and y = \frac{2 x}{7}$ $Rightarrow y = frac(- 5 x - 2)(8) and y = frac(2 x)(7)$

Then, let's set the two expressions for $y$ $y$ equal to each other:

$\Rightarrow \frac{- 5 x - 2}{8} = \frac{2 x}{7}$ $Rightarrow frac(- 5 x - 2)(8) = frac(2 x)(7)$

$\Rightarrow 7 \times (- 5 x - 2) = 8 \times 2 x$ $Rightarrow 7 times (- 5 x - 2) = 8 times 2 x$

$\Rightarrow - 35 x - 14 = 4 x$ $Rightarrow - 35 x - 14 = 4 x$

$\Rightarrow - 35 x - 4 x - 14 = 0$ $Rightarrow - 35 x - 4 x - 14 = 0$

$\Rightarrow - 39 x - 14 = 0$ $Rightarrow - 39 x - 14 = 0$

$\Rightarrow - 39 x = 14$ $Rightarrow - 39 x = 14$

$therefore x = - frac(14)(39)$

Now that we have a value of $x$ , let's find $y$ using one of the expressions:

$Rightarrow y = frac(2 x)(7)$

$Rightarrow y = frac(2 times - frac(14)(39))(7)$

$Rightarrow y = frac(- frac(28)(39))(7)$

$Rightarrow y = frac(- frac(28)(39))(frac(7)(1))$

$Rightarrow y = - frac(28)(39) times frac(1)(7)$

$Rightarrow y = - frac(28 times 1)(39 times 7)$

$Rightarrow y = - frac(28)(273)$

$therefore y = - frac(4)(39)$

Therefore, the solutions to the equation are $x = - frac(14)(39)$ and $y = - frac(4)(39)$ .

How do you solve the following system: 5x + 8y = -2, 4x + 7y = 6x5x+8y=−2,4x+7y=6x5x + 8y = -2, 4x + 7y = 6x?