Please help me with this

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Step-by-step explanation:

\begin{gathered}\frac{a}{x-b} +\frac{b}{x-a}=2 \:\:\:(*(x-a)(x-b))\\a(x-a) + b(x-b) =2 (x-a)(x-b)\\ax - a^2 + bx - b^2 = 2 (x-a) (x-b) \\ax - a^2+ bx - b^2 = (2x-2a) (x-b) \\ax - a^2 + bx - b^2 = 2x^2 - 2bx - 2ax + 2ab \\-2x^2 + 3bx + 3ax - a^2 - b^2 - 2ab = 0 \\2x^2 - 3bx - 3ax + a^2 + b^2 +2ab = 0 \\2x^2 - 3 (a+b) x + (a+b)^2 = 0 \\2x^2 - 2(a+b)x - 1(a+b)x + (a+b)^2 = 0 \\2x [x - (a+b)] - (a+b) [x - (a+b)] = 0 \\(2x - a - b ) (x - a - b) = 0 \\2x - a - b = 0\:\:\:\:OR\:\:\:\:x - a - b = 0\end{gathered}
x?b
a
?
+
x?a
b
?
=2(?(x?a)(x?b))
a(x?a)+b(x?b)=2(x?a)(x?b)
ax?a
2
+bx?b
2
=2(x?a)(x?b)
ax?a
2
+bx?b
2
=(2x?2a)(x?b)
ax?a
2
+bx?b
2
=2x
2
?2bx?2ax+2ab
?2x
2
+3bx+3ax?a
2
?b
2
?2ab=0
2x
2
?3bx?3ax+a
2
+b
2
+2ab=0
2x
2
?3(a+b)x+(a+b)
2
=0
2x
2
?2(a+b)x?1(a+b)x+(a+b)
2
=0
2x[x?(a+b)]?(a+b)[x?(a+b)]=0
(2x?a?b)(x?a?b)=0
2x?a?b=0ORx?a?b=0
?


Therefore,

x = a+b / 2\:\:\:OR\:\:\:x = a + bx=a+b/2 OR x=a+b

Hope this helps!

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