Question

Систему уравнений и решите сложения: a){x/2-y/3=1 {x/4+2y/3=8 б){x/4+y/4=2 {x/6+y/3=2 в){2x+x-y/4=11 {3y-x+y/3=1

Answer 1

А) $\left \{ {{ \frac{x}{2}- \frac{y}{3}=1 } \atop { \frac{x}{4}+ \frac{2y}{3}=8 }} \right. ; =\ \textgreater \ \left \{ {{3x-2y=2*3} \atop {3x+4*2y=8*3*4}} \right.; =\ \textgreater \ \left \{ {{3x-2y=6} \atop {3x+8y=96}} \right. .$
3x - 2y + 3x + 8y= 6 + 96
6х + 6у = 102
6(х + у) = 102
х + у = 102 : 6
х + у = 17
х = 17 - у

3(17 - у) - 2у = 6
51 - 3у - 2у = 6
51 - 5у = 6
5у = 51 - 6
5у = 45
у = 45 : 5
у = 9

х = 17 - 9
х = 8

ответ: х = 8; у = 9.

б) $\left \{ {{ \frac{x}{4}+ \frac{y}{4}=2 } \atop { \frac{x}{6} + \frac{y}{3} =2}} \right. ;=\ \textgreater \ \left \{ {{4x+4y=2*4*4} \atop {3x+6y=2*6*3}} \right. ;=\ \textgreater \ \left \{ {{4x+4y=32} \atop {3x+6y=36}} \right.;=\ \textgreater \ \left \{ {{x+y=8} \atop {x+2y=12}} \right. \\$

х + 2у - х - у = 12 - 8
у = 4

х + 4 = 8
х = 8 - 4
х = 4
ответ: х = 4; у = 4.

в)