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

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

1 Answer

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Answer 1 · 2016-01-11T15:13:47

$y = - 1$ $y=-1$
$x = \frac{1}{2}$ $x = 1/2$

Explanation:

Noting that the coefficient of $x$ $x$ is the same in both equations, we can get an expression just in $y$ $y$ buy subtracting the second equation from the first.
$4 x + 3 y = - 1$ $4x+3y = -1$
$- (4 x - 2 y = 4)$ $-(4x -2y = 4)$
$5 y = - 5$ $5y = -5$
$y = - 1$ $y=-1$
Substituting into either of the equations(I choose the second one) we can evaluate $x$ $x$
$4 x = 4 + 2 \cdot (- 1)$ $4x = 4+2*(-1)$
$4 x = 2$ $4x = 2$
$x = \frac{1}{2}$ $x = 1/2$

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

1 Answer

Explanation:

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How do you solve the following system: 4x+3y= -1 , 4x-2y=4 4x+3y=−1,4x−2y=44x+3y= -1 , 4x-2y=4 ?

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Explanation:

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