How do you solve the following system?: $y = .5 x + 2, 2 y = 4 x + 3$ $y= .5x+2 , 2y=4x+3$

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Answer 1 · 2016-02-13T10:39:58

Solution is $x = \frac{1}{3}, y = \frac{13}{6}$ $x=1/3,y=13/6$

The equations are $y = 0.5 x + 2$ $y=0.5x+2$ and $2 y = 4 x + 3$ $2y=4x+3$

Multiplying the first equation by $2$ $2$ we get

$2 y = x + 4$ $2y=x+4$ but $2 y = 4 x + 3$ $2y=4x+3$

Hence $x + 4 = 4 x + 3$ $x+4=4x+3$ or $3 x = 1$ $3x=1$ or $x = \frac{1}{3}$ $x=1/3$

Putting $x = \frac{1}{3}$ $x=1/3$ in $y = 0.5 x + 2$ $y=0.5x+2$

we get $y = 0.5 \cdot \frac{1}{3} + 2 = \frac{1}{6} + 2 = \frac{13}{6}$ $y=0.5*1/3+2=1/6+2=13/6$

Hence solution is $x = \frac{1}{3}, y = \frac{13}{6}$ $x=1/3,y=13/6$

How do you solve the following system?: y= .5x+2 , 2y=4x+3 y=.5x+2,2y=4x+3y= .5x+2 , 2y=4x+3