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Question Number 57698 by rahul 19 last updated on 10/Apr/19

Commented by rahul 19 last updated on 10/Apr/19

$$3rdplanecanbewrittenas:$$$$P_1+\mathrm{{\rm Y}}{P}_2,howtofind\mathrm{{\rm Y}}?$$$$here\lambda =3,\mu =7butthen\mathrm{{\rm Y}}isnotunique!$$

Answered by mr W last updated on 10/Apr/19

$$\left(ii\right)-\left(i\right):$$$$y+z=1...\left(iv\right)$$$$\left(iii\right)-\left(i\right):$$$$2y+(\lambda -1)z=\mu -5...\left(v\right)$$$$\left(v\right)-2\left(iv\right):$$$$\Rightarrow (\lambda -3)z=\mu -7$$$$\Rightarrow \lambda =3,\mu =7$$$$\lambda +\mu =10$$$$\Rightarrow \left(2\right)iscorrect.$$

Commented by rahul 19 last updated on 10/Apr/19

Sir, what's the problem in my method?

Commented by mr W last updated on 10/Apr/19

$$methodisoksir.youcangetthesame$$$$result.$$$$P_1+\mathrm{{\rm Y}}{P}_2:$$$$(1+\mathrm{{\rm Y}})x+(1+2\mathrm{{\rm Y}})y+(1+2\mathrm{{\rm Y}})z=(5+6\mathrm{{\rm Y}})$$$$\Rightarrow x+\frac{1+2\mathrm{{\rm Y}}}{1+\mathrm{{\rm Y}}}y+\frac{1+2\mathrm{{\rm Y}}}{1+\mathrm{{\rm Y}}}z=\frac{5+6\mathrm{{\rm Y}}}{1+\mathrm{{\rm Y}}}$$$$P_3:x+3y+\lambda z=\mu$$$$\Rightarrow \frac{1+2\mathrm{{\rm Y}}}{1+\mathrm{{\rm Y}}}=3\Rightarrow \mathrm{{\rm Y}}=-2$$$$\Rightarrow \frac{1+2\mathrm{{\rm Y}}}{1+\mathrm{{\rm Y}}}=\lambda =3$$$$\Rightarrow \frac{5+6\mathrm{{\rm Y}}}{1+\mathrm{{\rm Y}}}=\mu =6-\frac{1}{1+\mathrm{{\rm Y}}}=7$$$$\Rightarrow \lambda +\mu =3+7=10$$

Commented by rahul 19 last updated on 11/Apr/19

$$ThanksSir.$$$$Iwasdoing1+\mathrm{{\rm Y}}=1,1+2\mathrm{{\rm Y}}=3i.e$$$$i{didn}^{\prime }tmakecoeff.ofxas1.$$

Commented by mr W last updated on 11/Apr/19

$$thecoefficientsmustnotbethesame,$$$$onlytheratiosbetweenthemmustbe$$$$thesame.$$

Commented by rahul 19 last updated on 11/Apr/19

$$OkSir!:)$$

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