Question-200242

Question Number 200242 by cherokeesay last updated on 16/Nov/23

Answered by cortano12 last updated on 17/Nov/23

(((ab+c)x)/(b−(c/a) )) −(((ab−c)x)/(b+(c/a))) = ((ab−c)/(b+(c/a))) −((ab+c)/(b−(c/a))) ((ax(ab+c))/(ab−c)) −((ax(ab−c))/(ab+c)) = ((a(ab−c))/(ab+c)) −((a(ab+c))/(ab−c)) ((x(ab+c))/(ab−c)) −((x(ab−c))/(ab+c)) = ((ab−c)/(ab+c))−((ab+c)/(ab−c)) (((ab+c)(x+1))/(ab−c)) = (((ab−c)(x+1))/(ab+c)) (ab+c)^2 (x+1)= (ab−c)^2 (x+1) (x+1){(ab+c)^2 −(ab−c)^2 } = 0 (x+1)(2ab)(2c)=0 x = −1

$$\:\frac{\left({ab}+{c}\right){x}}{{b}−\frac{{c}}{{a}}\:}\:−\frac{\left({ab}−{c}\right){x}}{{b}+\frac{{c}}{{a}}}\:=\:\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}\:−\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}} \\ $$$$\:\frac{{ax}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{ax}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{a}\left({ab}−{c}\right)}{{ab}+{c}}\:−\frac{{a}\left({ab}+{c}\right)}{{ab}−{c}} \\ $$$$\:\:\frac{{x}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{x}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{ab}−{c}}{{ab}+{c}}−\frac{{ab}+{c}}{{ab}−{c}} \\ $$$$\:\frac{\left({ab}+{c}\right)\left({x}+\mathrm{1}\right)}{{ab}−{c}}\:=\:\frac{\left({ab}−{c}\right)\left({x}+\mathrm{1}\right)}{{ab}+{c}} \\ $$$$\:\left({ab}+{c}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)=\:\left({ab}−{c}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right) \\ $$$$\:\left({x}+\mathrm{1}\right)\left\{\left({ab}+{c}\right)^{\mathrm{2}} −\left({ab}−{c}\right)^{\mathrm{2}} \right\}\:=\:\mathrm{0} \\ $$$$\:\left({x}+\mathrm{1}\right)\left(\mathrm{2}{ab}\right)\left(\mathrm{2}{c}\right)=\mathrm{0} \\ $$$$\:{x}\:=\:−\mathrm{1} \\ $$

Commented by cherokeesay last updated on 16/Nov/23

$${thanks}\:! \\ $$

Answered by som(math1967) last updated on 16/Nov/23

((ax(ab+c))/(ab−c)) +((a(ab+c))/(ab−c))=((ax(ab−c))/(ab+c))+((a(ab−c))/(ab+c)) ⇒ax{(((ab+c)^2 −(ab−c)^2 )/(a^2 b^2 −c^2 ))}=a{(((ab−c)^2 −(ab+c)^2 )/(a^2 b^2 −c^2 ))} x=((−a)/a)=−1

$$\frac{{ax}\left({ab}+{c}\right)}{{ab}−{c}}\:+\frac{{a}\left({ab}+{c}\right)}{{ab}−{c}}=\frac{{ax}\left({ab}−{c}\right)}{{ab}+{c}}+\frac{{a}\left({ab}−{c}\right)}{{ab}+{c}} \\ $$$$\Rightarrow{ax}\left\{\frac{\left({ab}+{c}\right)^{\mathrm{2}} −\left({ab}−{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\right\}={a}\left\{\frac{\left({ab}−{c}\right)^{\mathrm{2}} −\left({ab}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\right\} \\ $$$${x}=\frac{−{a}}{{a}}=−\mathrm{1}\: \\ $$

Commented by cherokeesay last updated on 16/Nov/23

$${thank}\:{you}\:! \\ $$

Answered by Rasheed.Sindhi last updated on 16/Nov/23

(((ab+c)x)/(b−(c/a)))−((ab−c)/(b+(c/a)))=(((ab−c)x)/(b+(c/a)))−((ab+c)/(b−(c/a))) (((ab+c)/(b−(c/a)))−((ab−c)/(b+(c/a))))x=((ab−c)/(b+(c/a)))−((ab+c)/(b−(c/a))) a(((b+(c/a))/(b−(c/a)))−((b−(c/a))/(b+(c/a))))x=−a(((b+(c/a))/(b−(c/a)))−((b−(c/a))/(b+(c/a)))) [a≠0] x=−1

$$\frac{\left({ab}+{c}\right){x}}{{b}−\frac{{c}}{{a}}}−\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}=\frac{\left({ab}−{c}\right){x}}{{b}+\frac{{c}}{{a}}}−\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}} \\ $$$$\left(\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}}−\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}\right){x}=\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}−\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}} \\ $$$$\cancel{{a}}\left(\frac{{b}+\frac{{c}}{{a}}}{{b}−\frac{{c}}{{a}}}−\frac{{b}−\frac{{c}}{{a}}}{{b}+\frac{{c}}{{a}}}\right){x}=−\cancel{{a}}\left(\frac{{b}+\frac{{c}}{{a}}}{{b}−\frac{{c}}{{a}}}−\frac{{b}−\frac{{c}}{{a}}}{{b}+\frac{{c}}{{a}}}\right)\:\:\:\left[{a}\neq\mathrm{0}\right] \\ $$$${x}=−\mathrm{1} \\ $$

Commented by cherokeesay last updated on 16/Nov/23

$${thank}\:{you}\:! \\ $$

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