How do you solve the following system: $2 x - 4 y = 6, y = \frac{3}{4} x + 32$ $2x-4y=6 , y= 3/4x+3 2$ ?

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Answer 1 · 2016-02-27T01:24:15

x=-134; y=137/2

$a) 2 x - 4 y = 6$ $a) 2x-4y=6$

$b) y = \frac{3}{4} x + 32$ $b) y=3/4x+32$

You just need to replace $y$ $y$ in $a)$ $a)$ by its value in $b$ $b$ :

$a) 2 x - 4 (\frac{3}{4} x + 32) = 6$ $a) 2x-4(3/4x+32)=6$

$a) 2 x - 3 x - 128 = 6$ $a) 2x-3x-128=6$

$a) - x = 6 + 128 = 134$ $a) -x=6+128=134$

$a) x = - 134$ $a) x=-134$

Now that we have the value of x we wil replace it in equation b:

$b) y = (\frac{3}{4}) \times (- 134) + 32 = - \frac{201}{2} + 32 = \frac{137}{2}$ $b) y=(3/4)xx(-134)+32=-201/2+32= 137/2$

How do you solve the following system: 2x-4y=6 , y= 3/4x+3 2 2x−4y=6,y=34x+322x-4y=6 , y= 3/4x+3 2 ?