. The young algebraist's companion : or, A new and easy guide to algebra; introduced by the doctrine of vulgar fractions: designed for the use of schools ... illustrated with variety of numerical and literal examples, and attempted in natural and familar dialogues ... r 2 3 4 Therefore, ^ The Square of ^ + * is hb -|- 2bx + A*; (Q^) is IZ Square of ;«• and a* Whence, bb + ^bx -\- xx IZ xx -{ aa. By cancelUng xx on both Sides, bb -f- 2iA zz ^^.4. Then 9 Z*^, it will be 2^a- zz ^^ — bb, aa—bb 12X 2 — 4X4 t zb 2X4 Sothatthe SideC^or C£) zz;f ZZ 16. AndCD i6-{-BD ZZ4ZZ20ZZ Hypothenufe CB. PR
. The young algebraist's companion : or, A new and easy guide to algebra; introduced by the doctrine of vulgar fractions: designed for the use of schools ... illustrated with variety of numerical and literal examples, and attempted in natural and familar dialogues ... r 2 3 4 Therefore, ^ The Square of ^ + * is hb -|- 2bx + A*; (Q^) is IZ Square of ;«• and a* Whence, bb + ^bx -\- xx IZ xx -{ aa. By cancelUng xx on both Sides, bb -f- 2iA zz ^^.4. Then 9 Z*^, it will be 2^a- zz ^^ — bb, aa—bb 12X 2 — 4X4 t zb 2X4 Sothatthe SideC^or C£) zz;f ZZ 16. AndCD i6-{-BD ZZ4ZZ20ZZ Hypothenufe CB. PROBLEM XLIIL There is a rcSfangled Triangled ABC, whofe Bafe ABzz 45, and the Su?n of the Hypothenufe and Ca^t bet us AC + BC :z: 135 : It is required to findthe Sides AC and QQ feparately F Literal Algebraic Problems. 207 Literal Solidio?:. Let ^/ z:: 135 = ^C -f BC, andlet b zz. 45, or BA^ and put x forCAy then CB — d — x. And fee-ing the Angle CAB is a right one,we have (by the 47^/^ of i Euc.)BO^ z= C^ + BJ1; but 5C^Zl the Square of d — ^, viz. dd —^dx + XX. And C^^ zz . j andjByf ~ ii. I 2 3 4 The Square d — ;^ is dd — idx -\- xx. This (Q:^) zz to the Square of x and , dd — 2dx •\ xx zz xx -\ cancelling xx^ dd — idx zz, bb. Then ^ bby dd — ^^ zz 2dx^ Therefore, X dd-bh d bb , — or, ; ~ OQ 2^, 2 2^ zz CA, And 135 — 60 = 75 = BC, 0 R, Let X r: -5C, and tlien will AC zz d — x ,therefore (by the 47/^ of i Euc.) BC^ ZZ AC- +AB^, Therefore, as ^ zz BA^ as above, it will be I2 3 XX zz dd — 2dx -^ XX -\- bbyo zz: dd -^ 2dx -\- bb^2dx zz dd -^ bb, -^ ~ zz 7 c the 2 X 135 ^^ Hypothenufe BC, And 135 — 75 zz 60 zz ACy as before. A* _^ bb _ IK , -2 ^^-T^ T 2 iV<;/^,. 2o8 Problems. Ncte^ By the 36/^ of 3 Euclid^ you may find the Sides BC and AC thus : Defcribe a Circle, makingthe Perpendicular the Radius ; then is the Re<Slanglc Be into the Segment BD ZZ the ,• -. Square of JB ; therefore BD is / .-^
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