How do you solve the system $\frac{1}{2} x + \frac{3}{4} y = - \frac{1}{4}$ $1/2x + 3/4y = -1/4$ and $\frac{3}{4} x + - \frac{1}{4} y = 1$ $3/4x + -1/4y = 1$ ?

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How do you solve the system $\frac{1}{2} x + \frac{3}{4} y = - \frac{1}{4}$ $1/2x + 3/4y = -1/4$ and $\frac{3}{4} x + - \frac{1}{4} y = 1$ $3/4x + -1/4y = 1$ ?

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Answer 1 · 2018-06-13T06:24:30

$x = 1$ $x=1$ and $y = - 1$ $y=-1$

Explanation:

$\frac{1}{2} x + \frac{3}{4} y = - \frac{1}{4}$ $1/2 x+3/4 y=-1/4$
Multiply all by 4
$2 x + 3 y = - 1$ $2x+3y=-1$

$\frac{3}{4} x + - \frac{1}{4} y = 1$ $3/4 x+ -1/4 y=1$
$\frac{3}{4} x - \frac{1}{4} y = 1$ $3/4x-1/4y=1$
Multiply all by 4
$3 x - y = 4$ $3x-y=4$
$y = 3 x - 4$ $y=3x-4$

Substitute $y$ $y$ into the equation
$2 x + 3 (3 x - 4) = - 1$ $2x+3(3x-4)=-1$
$2 x + 9 x - 12 = - 1$ $2x+9x-12=-1$
$11 x = 11$ $11x=11$
$x = 1$ $x=1$

Substitute $x$ $x$ into the equation
$y = 3 (1) - 4$ $y=3(1)-4$
$y = 3 - 4$ $y=3-4$
$y = - 1$ $y=-1$

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How do you solve the system $\frac{1}{2} x + \frac{3}{4} y = - \frac{1}{4}$ $1/2x + 3/4y = -1/4$ and $\frac{3}{4} x + - \frac{1}{4} y = 1$ $3/4x + -1/4y = 1$ ?

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How do you solve the system $\frac{1}{2} x + \frac{3}{4} y = - \frac{1}{4}$ $1/2x + 3/4y = -1/4$ and $\frac{3}{4} x + - \frac{1}{4} y = 1$ $3/4x + -1/4y = 1$ ?