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

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Answer 1 · 2017-05-10T11:07:07

Instead of y in the second question, write the first equation (of the right hand side) to solve

$3 x - 3 y = 2$ $3x-3y=2$

Instead of y in the above equation, write ( $3 \frac{x}{4} + 3$ $3x/4+3$ ) where you see y.

It is:

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

$3 x - 9 \frac{x}{4} - 9 = 2$ $3x-9x/4-9=2$

$\frac{12 x - 9 x}{4} = 11$ $(12x-9x)/4=11$

$3 \frac{x}{4} = 11$ $3x/4=11$

$x = \frac{44}{3}$ $x=44/3$

From here you can solve y:

$y = (\frac{3}{4}) \cdot (\frac{44}{3}) + 3$ $y=(3/4)*(44/3) + 3$

$y = 11 + 3$ $y=11+3$

$y = 14$ $y=14$

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