a/(x-a)+b/(x-b)=2c/(x-c)
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a/x-a+b/x-b=2c/x-c
=>a/(x-a)-c/(x-c)=c/(x-c)-b/(x-b)
=>ax-ac-cx+ac/(x-a)(x-b)=cx-bc-bx+bc/(x-c)(x-b)
=>(ax-cx)/(x-a)(x-c)=(cx-bx)/(x-b)(x-c)
=>(ax_cx)/(x-a)(x-c)-(cx-bx)/(x-b)(x-c)=0
=>1/(x-c){(ax-cx)/(x-a)-(cx-bx)/(x-b)}=0
so,
1/(x-c)=0
=>x-c=0
=>x=c
or,
{(ax-cx)/(x-a)-(cx-bx)/(x-b)}=0
=>{(ax-cx)/(x-a)}={(cx-bx)/(x-b)}
from here u can solve the value of x
please mark as brainliest answer if you get help from this
=>a/(x-a)-c/(x-c)=c/(x-c)-b/(x-b)
=>ax-ac-cx+ac/(x-a)(x-b)=cx-bc-bx+bc/(x-c)(x-b)
=>(ax-cx)/(x-a)(x-c)=(cx-bx)/(x-b)(x-c)
=>(ax_cx)/(x-a)(x-c)-(cx-bx)/(x-b)(x-c)=0
=>1/(x-c){(ax-cx)/(x-a)-(cx-bx)/(x-b)}=0
so,
1/(x-c)=0
=>x-c=0
=>x=c
or,
{(ax-cx)/(x-a)-(cx-bx)/(x-b)}=0
=>{(ax-cx)/(x-a)}={(cx-bx)/(x-b)}
from here u can solve the value of x
please mark as brainliest answer if you get help from this
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