Решите систему уравенеий 4х-3у=5 3х+4у=10

Уравняем коэффициенты перед х, для чего,первое уравнение системы домножим на 3, а вторую на 4, получим: 12х-9у=15 12х+16у=40  из второго уравнения почленно вычтем первое: 25у=25 у=1 подставим у=1 в первое уравнение системы 12х-9=15                                                                                   12х=24                                                                                         х=2 ответ (2; 1) выразим х из первого уравнения х=(5+3у)/4, подставим х во второе уравнение системы: 3*(5+3у)/4 + 4у =10 домножим обу части уравнения на 4 : 3*(5 + 3у) +16 у=40 15 + 9у + 16у = 40 25у=25 у=1 подставим у=1 в первое уравнение системы 4х-3=5                                                                                   4х=8                                                                                         х=2

Можно решить так: умножим первое ур. на 4 . а второе ур. на 3 получим 16х-12у=20 9х+12у=30 сложим получившиеся уравнения  :   25х=50 откуда х= 2 подставляем значения х в любое ур. 4* 2 -3у=5 8-3у=5 3у=3 у=1

Решение.

Впишем четырехугольник ABCD в прямоугольник EFGH со сторонами,

параллельными диагоналям (EF || AC и EH || BD) - смотри рисунок.

Пусть L - точка пересечения прямых DC и EF, а M - точка на прямой HG такая, что LM || FG.

Тогда ABLC - параллелограмм, следовательно, AB = CL.

Так как GM = FL = EB = HD и AH = CG, то треуг-к AHD = треуг-ку CGM ,

следовательно, AD = CM. BC + CM = BC + AD .

Но BM = DL как диагонали прямоугольника BLDM, и DL = DC + CL = DC + AB.

Следовательно, AD + BC = DL = DC + CL = DC + AB, что и требовалось доказать.

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