SOLUTION: Hi there, how can I find an equation of th line containing (2,-1) and parallel to y=1/3� + 2? I need help solving for u: 5u^2-50=0 How can I solve for x? 2��+2=5�2�-1

Question 818866: Hi there, how can I find an equation of th line containing (2,-1) and parallel to y=1/3� + 2?

I need help solving for u: 5u^2-50=0

How can I solve for x? 2��+2=5�2�-1
Rationalizing denominator: -5�3 square root 2 + 6

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Hi there, how can I find an equation of th line containing (2,-1) and parallel to y=1/3� + 2?
--------
Form: y = mx + b
m = (1/3)
x = 2
y = -1
Solve for "b":
-1 = 2*2 + b
b = -5
----
Equation:
y = (1/3)x - 5
============================
I need help solving for u:
5u^2-50=0
Divide thru by 5 to get:
u^2 - 10 = 0
u^2 = 10
u = sqrt(10) or u = -sqrt(10)
===============================
==================================
How can I solve for x? 2��+2=5�2�-1
----
2/(x+2) = 5/(2x-1)
Cross-multiply to get:
4x-2 = 5x+10
x = -12
-----------------
Rationalizing denominator: -5�3 square root 2 + 6
-5/(3sqrt(2) + 6)
-----
Multiply numerator and denominator by 3sqrt(2)-6 to get:
[-5(3sqrt(2)-6)]/[(3sqrt(2)+6)(3sqrt(2)-6)]
---------
= [-15sqrt(2)+30]/[(3sqrt(2))^2-6^2]
-----
[-15sqrt(2)-30] / [9*2-36]
----
= (-15sqrt(2)-30) / (-18)
----
= (5/6)sqrt(2)+ (5/3)
===========================
Cheers,
Stan H.
==================

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