Algebra Pre-Algebra and Basic Algebra Math Forum

 May 31st, 2014, 12:14 AM #1 Newbie   Joined: May 2014 From: bangkok Posts: 4 Thanks: 0 Help please How do you solve x+1 over x+3 =2x-1 over 3x + 1 Thanks
 May 31st, 2014, 02:22 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Hint: if a/b = c/d, then ad = bc
 May 31st, 2014, 06:18 PM #3 Newbie   Joined: May 2014 From: Bangkok Posts: 6 Thanks: 0 A little extra x+1 over x+3 =2x-1 over 3x + 1 so, (x+1)(3x+1)(x+3)(2x-1)? Sorry, to sound stupid I just can't expand it
 May 31st, 2014, 07:58 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Are you posting under different names?
 June 1st, 2014, 03:53 AM #5 Senior Member   Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 $\displaystyle \frac { x+1 }{ x+3 } =\frac { 2x-1 }{ 3x+1 } \\ \\ \frac { x+1 }{ \left( x+3 \right) } -\frac { \left( 2x-1 \right) }{ \left( 3x+1 \right) } =0\\ \\ \frac { \left( x+1 \right) \left( 3x+1 \right) }{ \left( x+3 \right) \left( 3x+1 \right) } -\frac { \left( 2x-1 \right) \left( x+3 \right) }{ \left( x+3 \right) \left( 3x+1 \right) } =0\\ \\ \frac { 3{ x }^{ 2 }+x+3x+1-\left[ 2{ x }^{ 2 }+6x-x-3 \right] }{ \left( x+3 \right) \left( 3x+1 \right) } =0\\ \\ 3{ x }^{ 2 }+4x+1-\left[ 2{ x }^{ 2 }+5x-3 \right] =0\\ \\ 3{ x }^{ 2 }+4x+1-2{ x }^{ 2 }-5x+3=0\\ \\ { x }^{ 2 }-x+4=0\\ \\ { x }^{ 2 }-x=-4\\ \\ { \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=-4\\ \\ { \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }=-\frac { 15 }{ 4 } \\ \\ \therefore \quad No\quad solution.\\ \\$ This can be seen on this graph: https://www.desmos.com/calculator/ogc8fwn2hl. Last edited by perfect_world; June 1st, 2014 at 03:56 AM.

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