Solve for x
1/a + 1/b + 1/x =1/a+b+x
Answers
Correct Question :
\displaystyle \text{{Solve for x : }}Solve for x :
\displaystyle \sf{{\frac{1}{a+b+x} =\frac{1}{a} +\frac{1}{b} +\frac{1}{x} }}
a+b+x
1
=
a
1
+
b
1
+
x
1
Answer:
\displaystyle \sf{{x=-a \ \ OR \ \ x=-b}}x=−a OR x=−b
Step-by-step explanation:
\displaystyle \sf{{Given : }}Given:
\displaystyle \sf{{\frac{1}{a+b+x} =\frac{1}{a} +\frac{1}{b} +\frac{1}{x} }}
a+b+x
1
=
a
1
+
b
1
+
x
1
\displaystyle \text{{Take $\frac{1}{\text{x}} $ L.H.S. side : }}Take
x
1
L.H.S. side :
$$\begin{lgathered}\displaystyle \sf{{\frac{1}{a+b+x}-\frac{1}{x} =\frac{1}{a} +\frac{1}{b} }}\\\\\\\displaystyle \sf{{\frac{x-(a+b+x)}{(a+b+x)(x)}=\frac{b+a}{ab} }}\end{lgathered}$$
$$\begin{lgathered}\displaystyle \sf{{\frac{x-a-b-x}{(a+b+x)(x)}=\frac{a+b}{ab} }}\\\\\\\displaystyle \sf{{\frac{-(a+b)}{(ax+bx+x^2)}=\frac{a+b}{ab} }}\end{lgathered}$$
$$\displaystyle \text{{Now (a + b) get cancel out : }}$$
$$\begin{lgathered}\displaystyle \sf{{\frac{-1}{(ax+bx+x^2)}=\frac{1}{ab} }}\\\\\\\displaystyle \sf{{ax+bx+x^2=-ab}}\\\\\\\displaystyle \sf{{x^2+ax+bx+ab=0}}\end{lgathered}$$
$$\displaystyle \text{{Take out 'x' and 'b' common : }}$$
$$\begin{lgathered}\displaystyle \sf{{x(x+a)+b(x+a)=0}}\\\\\\\displaystyle \sf{{(x+a)(x+b)}}\end{lgathered}$$
$$\begin{lgathered}\displaystyle \sf{{\rightarrow x+a=0}}\\\\\\\displaystyle \sf{{\rightarrow x = -a}}\\\\\\\displaystyle \sf{{OR}}\\\\\\\displaystyle \sf{{\rightarrow x+b=0}}\\\\\\\displaystyle \sf{{\rightarrow x=-b}}\end{lgathered}$$
$$\displaystyle \text{{Therefore , the value of x is '-a' OR ' -b' .}}$$
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