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Question Number 97454 by bemath last updated on 08/Jun/20

Answered by john santu last updated on 08/Jun/20

$$\left(1\right)17\mathrm{x}+51\mathrm{y}=85[:17]$$$$\Rightarrow \mathrm{x}+3\mathrm{y}=5\Rightarrow 19\mathrm{x}+57\mathrm{y}=19\times 5=95$$$$\left(2\right)\sqrt{10-\mathrm{x}}=6-\sqrt{4+\mathrm{x}}\left(\mathrm{square}\right)$$$$15+\mathrm{x}=6\sqrt{4+\mathrm{x}}\left[\mathrm{square}\right]$$$${\mathrm{x}}^2-6\mathrm{x}+81=0.\mathrm{Then}$$$$\sqrt{(4+\mathrm{x})(10-\mathrm{x})}=\sqrt{121-({\mathrm{x}}^2-6\mathrm{x}+81)}=11$$$$\left(3\right)\mathrm{the}\mathrm{first}\mathrm{inequality}\mathrm{is}\mathrm{equivalent}$$$$\mathrm{to}{\mathrm{x}}^2-2\mathrm{x}-8⩽0,\mathrm{while}\mathrm{the}\mathrm{second}$$$$\mathrm{is}\mathrm{equivalent}\mathrm{to}{\mathrm{x}}^2-2\mathrm{x}-8⩾0$$$$\mathrm{since}\mathrm{their}\mathrm{intersection}\mathrm{is}$$$${\mathrm{x}}^2-2\mathrm{x}-8=0,\mathrm{or}(\mathrm{x}+2)(\mathrm{x}-4)=0$$$$\mathrm{it}\mathrm{follows}\mathrm{that}\mathrm{x}=-2;4$$$$\left(4\right)\mathrm{Let}a\mathrm{and}b\mathrm{represent}\mathrm{the}$$$$\mathrm{length}\mathrm{and}\mathrm{width}\mathrm{of}\mathrm{rectangle}$$$$\mathrm{Then}2a^2+2b^2=100\mathrm{or}{a}^2+b^2=50$$$$\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{diagonal}\mathrm{is}\sqrt{a^2+b^2}=5\sqrt{2}$$$$\left(5\right)\mathrm{since}\mathrm{both}\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{are}\mathrm{positive},$$$$\log {}_2\left(\mathrm{p}\right)+\log {}_2\left(\mathrm{q}\right)=\log {}_2\left(\mathrm{pq}\right)$$$$\mathrm{but}\mathrm{p}\mathrm{\&}\mathrm{q}\mathrm{are}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}$$$$\mathrm{equation}{2\mathrm{x}}^2-5\mathrm{x}+1=0,\mathrm{so}$$$$\mathrm{pq}=\frac{1}{2},\mathrm{such}\mathrm{that}\log {}_2\left(\mathrm{pq}\right)=\log {}_2\left(\frac{1}{2}\right)=-1$$

Commented by bemath last updated on 08/Jun/20

coolll

Answered by som(math1967) last updated on 08/Jun/20

$$1.17\mathrm{x}+51\mathrm{y}=85$$$$\Rightarrow 17(\mathrm{x}+3\mathrm{y})=85$$$$\Rightarrow \mathrm{x}+3\mathrm{y}=5$$$$\Rightarrow 19\mathrm{x}+57\mathrm{y}=95\mathrm{ans}$$

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