How do you solve the following system?: $6 x - 4 y = 3, 5 x - 2 y = 1$ $6x - 4y = 3 , 5x - 2y = 1$

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How do you solve the following system?: $6 x - 4 y = 3, 5 x - 2 y = 1$ $6x - 4y = 3 , 5x - 2y = 1$

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Answer 1 · 2015-12-02T11:14:59

$x = \frac{- 1}{4}$ $x=(-1)/4$
$y = \frac{- 9}{8}$ $y=(-9)/8$

Explanation:

$6 x - 4 y = 3$ $6x-4y=3$ --------------(1)
$5 x - 2 y = 1$ $5x-2y=1$ ------------------(2)
Let us have the equations as -

We are going to eliminate $y$ $y$

Before that understand $- 4 y + 4 y = 0$ $-4y +4y =0$

Multiply equation (2) with $- 2$ $-2$

$6 x - 4 y = 3$ $6x-4y=3$ ------------------(1)
$- 10 x + 4 y = - 2$ $-10x+4y=-2$ ------------------(2)

Add (1) and (2), we get -

$- 4 x = 1$ $-4x=1$
$x = \frac{- 1}{4}$ $x=(-1)/4$

Substitute $x = \frac{- 1}{4}$ $x=(-1)/4$ in equation (1)

$6 (\frac{- 1}{4}) - 4 y = 3$ $6((-1)/4)-4y=3$
$\frac{- 6}{4} - 4 y = 3$ $(-6)/4-4y=3$
$\frac{- 3}{2} - 4 y = 3$ $(-3)/2-4y=3$
$- 4 y = 3 + \frac{3}{2} = \frac{6 + 3}{2} = \frac{9}{2}$ $-4y = 3+3/2=(6+3)/2=9/2$
$y = \frac{9}{2} \times \frac{1}{- 4} = \frac{- 9}{8}$ $y=9/2 xx1/(-4)=(-9)/8$

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How do you solve the following system?: $6 x - 4 y = 3, 5 x - 2 y = 1$ $6x - 4y = 3 , 5x - 2y = 1$

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Explanation:

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