Question

Решить системы уравнений:
1) x - 2y = 4
3x + 5y = 10
2) 7x + 3y = 2
2x - 15y = 1
3) x - 7y = 2
3y - 2x = 8

Answer 1

$1) ; left{begin{array}{ccc}x-2y=4\3x+5y=10end{array}right;Longrightarrow;;;left{begin{array}{ccc}x=4+2y\3(4+2y)+5y=10end{array}right\\3(4+2y)+5y=10\12+6y+5y=10\12+11y=10\11y=-2\y=-2:11\y=-dfrac2{11}\\\x=4+2y\x=4+2cdot(-dfrac2{11})\x=4-dfrac{4}{11}\\x=3dfrac{7}{11}\\x=dfrac{40}{11}\\\Ombeta em: ; (dfrac{40}{11};;-dfrac{2}{11})$

$2);left{begin{array}{ccc}7x+3y=2\2x-15y=1end{array}rightRightarrowleft{begin{array}{ccc}3y=2-7x\2x-15y=1end{array}rightRightarrowleft{begin{array}{ccc}y=dfrac23-dfrac73x\\2x-15cdot(dfrac23-dfrac73x)=1end{array}right$

$2x-15cdot(dfrac23-dfrac73x)=1\\2x-dfrac{30}3+dfrac{105}{3}x=1\\2x-10+35x=1\37x-10=1\37x=11\\x=dfrac{11}{37}\\\y=dfrac23-dfrac73cdotdfrac{11}{37}\\y=dfrac23-dfrac{77}{111}\\y=dfrac{74}{111}-dfrac{77}{111}\\y=-dfrac{3}{111}\\y=-dfrac{1}{37}\\Ombeta em:;(dfrac{11}{37};;-dfrac{1}{37})$

$left{begin{array}{ccc}x-7y=2\3y-2x=8end{array}right;Longrightarrow;;;left{begin{array}{ccc}x=2+7y\3y-2(2+7y)=8end{array}right\\3y-2(2+7y)=8\3y-4-14y=8\-11y-4=8\11y+4=-8\11y=-12\y=-dfrac{12}{11}\\\x=2+7(-dfrac{12}{11})\\x=2-dfrac{84}{11}\\x=dfrac{22}{11}-dfrac{84}{11}\\x=-dfrac{62}{11}\\\Ombeta em:;(-dfrac{62}{11};;-dfrac{12}{11})$