Question

Bµi 1: A)\(\left\{{}\begin{matrix}X=35.\left(Y+2\right)\\X=50.\left(Y-1\right)\end{matrix}\right.\)

b)\(\left\{{}\begin{matrix}Y=2X-3\\Y=X-1\end{matrix}\right.\)

C) \(\left\{{}\begin{matrix}\left(X+14\right).\left(Y-2\right)=X.Y\\\left(X-4\right).\left(Y+1\right)=X.Y\end{matrix}\right.\)

D)\(\left\{{}\begin{matrix}Y=\frac{6-X}{4}\\Y=\frac{4X-5}{3}\end{matrix}\right.\)GIẢI BÀI 1 BẰNG PHƯƠNG PHAP THẾ

Answer 1

a,\(\left\{{}\begin{matrix}x=35\left(y+2\right)\\x=50\left(y-1\right)\end{matrix}\right.\)

suy ra :35(y+2)=50(y-1)

=>35y+70=50y-50

=>y=8

=>x=350

vậy :\(\left\{{}\begin{matrix}x=350\\y=8\end{matrix}\right.\)

b.\(\left\{{}\begin{matrix}y=2x-3\\y=x-1\end{matrix}\right.\)

suy ra: 2x-3=x-1

=>x=2

=>y=1

vậy \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

c.\(\left\{{}\begin{matrix}\left(x+14\right).\left(y-2\right)=xy\\\left(x-4\right).\left(y-1\right)=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-2x+14=0\\-x-y=0\end{matrix}\right.\)

vậy:\(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d,\(\left\{{}\begin{matrix}y=\frac{6-x}{4}\\y=\frac{4x-5}{3}\end{matrix}\right.\)

x=2

y=1

vậy...