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Question Number 145801 by mathdanisur last updated on 08/Jul/21

	Answered by ajfour last updated on 08/Jul/21
	$xy+tz=a solving for x,y in terms of t,z from 2nd and 3rd eqs. x=((bz−ct)/(z^2 −t^2 )) ; y=((bt−cz)/(t^2 −z^2 )) (((bz−ct)(cz−bt))/((z^2 −t^2 )^2 ))+tz=a ..(i) x+y+z+t=((b+c)/(z+t))+z+t=Q if at all z≠t ((bc(z+t)^2 −(b+c)^2 tz)/((z+t)^2 {(z+t)^2 −4tz}))+tz=a ......$
	$x y + t z = a$ $s o l v i n g f o r x, y i n t e r m s o f$ $t, z f r o m 2 n d a n d 3 r d e q s .$ $x = \frac{b z - c t}{z^{2} - t^{2}}; y = \frac{b t - c z}{t^{2} - z^{2}}$ $\frac{(b z - c t) (c z - b t)}{{(z^{2} - t^{2})}^{2}} + t z = a . . (i)$ $x + y + z + t = \frac{b + c}{z + t} + z + t = Q$ $i f a t a l l z \neq t$ $\frac{b c {(z + t)}^{2} - {(b + c)}^{2} t z}{{(z + t)}^{2} {{(z + t)}^{2} - 4 t z}} + t z = a$ $. . . . . .$
	Commented by mathdanisur last updated on 08/Jul/21

	$T h a n k y o u S e r$
	Answered by Rasheed.Sindhi last updated on 08/Jul/21
	${ ((xy+zt=38.............(i))),((xz+yt=34.............(ii))),((xt+yz=43.............(iii))) :} (ii)+(iii)⇒ (x+y)(z+t)=77 77=1×77=7×11 { ((x+y=1∧z+t=77⇒x+y+z+t=78)),((x+y=7∧z+t=11⇒x+y+z+t=18)) :} Possible values for x+y+z+t {18,78} (iii)+(i):(x+z)(y+t)=81 81=1×81=3×27=9×9 Possible values for x+y+z+t {82,30,18} (i)+(ii)⇒(x+t)(y+z)=72 72=1×72=2×36=3×24=4×18 =6×12=8×9 Possible values for x+y+z+t {73,38,27,22,18,17} In all the three cases 18 is common. • •^(•) x+y+z+t=18$
	${\begin{cases} x y + z t = 38 . . . . . . . . . . . . . (i) \\ x z + y t = 34 . . . . . . . . . . . . . (i i) \\ x t + y z = 43 . . . . . . . . . . . . . (i i i) \end{cases}$ $(i i) + (i i i) \Rightarrow (x + y) (z + t) = 77$ $77 = 1 \times 77 = 7 \times 11$ ${\begin{cases} x + y = 1 \land z + t = 77 \Rightarrow x + y + z + t = 78 \\ x + y = 7 \land z + t = 11 \Rightarrow x + y + z + t = 18 \end{cases}$ $P o s s i b l e v a l u e s f o r x + y + z + t$ ${18, 78}$ $(i i i) + (i) : (x + z) (y + t) = 81$ $81 = 1 \times 81 = 3 \times 27 = 9 \times 9$ $P o s s i b l e v a l u e s f o r x + y + z + t$ ${82, 30, 18}$ $(i) + (i i) \Rightarrow (x + t) (y + z) = 72$ $72 = 1 \times 72 = 2 \times 36 = 3 \times 24 = 4 \times 18$ $= 6 \times 12 = 8 \times 9$ $P o s s i b l e v a l u e s f o r x + y + z + t$ ${73, 38, 27, 22, 18, 17}$ $I n a l l t h e t h r e e c a s e s 18 i s c o m m o n .$ $\overset{∙}{∙ ∙} x + y + z + t = 18$
	Commented by Rasheed.Sindhi last updated on 08/Jul/21

	$T h e a n s w e r h a s b e e n r e v i s e d n o w .$
	Commented by mathdanisur last updated on 08/Jul/21

	$T h a n k y o u S e r$
	Commented by mathdanisur last updated on 08/Jul/21

	$a n s w e r : 18 S e r$
	Commented by Rasheed.Sindhi last updated on 08/Jul/21

	$S o r r y I {d i d n}^{'} t n o t i c e :$ $x, y, z, t \in Z^{+}$ $T w o a n s w e r s 18 & - 18 a r e$ $i n c a s e x, y, z, t \in Z$
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