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August 22nd, 2010, 02:45 PM  #1 
Newbie Joined: Aug 2010 Posts: 3 Thanks: 0  These equations are driving me to insanity!
Here is the first one: x2 ?5x =0 Looking at this equation, I assumed it was a quadratic because there is an = 0 sign at the end. But quadratics have the form 'xsquared + bx + c = 0'. I cannot find a solution to this question. Here is the other one: x +3 ? 1/x = 0 Same thing here  absolutely baffled. Somebody please help  I'm loading my shotgun as we speak! 
August 22nd, 2010, 02:53 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,823 Thanks: 1049 Math Focus: Elementary mathematics and beyond  Re: These equations are driving me to insanity!
x^2  5x = 0, x(x  5) = 0, x = 5, x = 0. x + 3  1/x = 0, multiply both sides by x, x^2 + 3x 1 = 0, x = (3 ± ?(13))/2. 
August 22nd, 2010, 03:35 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs  Re: These equations are driving me to insanity!
You were right, the first one is quadratic, c = 0. In the general form ax˛ + bx + c "b" and "c" can be zero, but "a" cannot. 
August 22nd, 2010, 05:42 PM  #4  
Newbie Joined: Aug 2010 Posts: 3 Thanks: 0  Re: These equations are driving me to insanity! Quote:
Bit new to the subject and was having trub getting my head round it. But to ask for your expertise once again. Can you please help with these ones. The first one is a simultaneous equation  (which i am still new to) and the second is asking me to express x in terms of y 2a  b = 8 3a + 2b = 5 And 'Write out a correct rearrangement of the formula to express x in terms of y' y =4(x +3) ?2x  
August 22nd, 2010, 05:52 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,823 Thanks: 1049 Math Focus: Elementary mathematics and beyond  Re: These equations are driving me to insanity!
First one: [1] 2a  b = 8 [2] 3a + 2b = 5 Multiply [1] by 2: [3] 4a  2b = 16 Add [2] and [3]: 7a = 21, a = 21/7 = 3. From [1]: 2(3)  b = 8, b = 6  8, b = 2. Second one: y = 4(x +3)  2x y = 4x + 12  2x y = 2x + 12 y  12 = 2x (y  12)/2 = x. 

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