Determine if the lines are parallel, perpendicular, or neither x-3y= 15 and y = -3+4

1 answer:

7 0

If the second equation is y=-3x+4 then they would be parallel and I found that by turning the first equation into y=mx+b and you subtract x from both sides of the equation and get -3y=-x+15 and then you divide both sides of the equation by -3 to get the y by itself and that turns into y=1/3x-5 and since 1/3 is the reciprocal of -3 they are perpendicular.

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Step-by-step explanation:

We need to solve the equation f(x)=g(x)

where $f(x)=(\frac{1}{2})^2-3$

and $g(x) = 3x + 5$

Solving to find f(x)=g(x)

Putting values of f(x) and g(x)

$f(x)=g(x)\\(\frac{1}{2})^2-3=3x + 5\\\frac{1}{4}-3=3x + 5\\Subtarct\,\,5\,\,on\,\,both\,\,sides:\\\frac{1}{4}-3-5=3x + 5-5\\\frac{1}{4}-8=3x\\Taking\,\,LCm\,\,on\,\,left\,\,side\\\frac{1-8*4}{4}=3x\\\frac{-31}{4}=3x\\Divide\,\,both\,\,sides\,\,by\,\,3\\x=\frac{-31}{4*3}\\x=\frac{-31}{12}$