Question

Rezolvati prin metoda substitutiei:$\left \{ {{( \sqrt{2}-1)x+( \sqrt{2}+1)y=6 \atop{( \sqrt{2}+1)x+ \sqrt{2}-1)y=2 }} \right.$

Answer 1

notam √2-1=a
notam √2+1=b
ax+by=6
bx+ay=2
x=(6-by)/a
b(6-by)/a+ay=2
6b-b²y+a²y=2a
6b-2a=y(b²-a²)
y=(6b-2a)/(b²-a²)=(6b-2a)/(b-a)(b+a)=
y=6(√2+1)-2(√2-1)/(√2+1-√2+1)(√2+1+√2-1)
y=(6√2+6-2√2+2)/4√2
y=(4√2+8)/4√2
y=1+√2
x=(6-by)/a
x=[6-(√2+1)(1+√2)]/(√2-1)
x=[6-(√2+1)²]/(√2-1)
x=[6-2-2√2-1)]/(√2-1)
x=(3-2√2)/(√2-1)    dar 3-2√2=2-2√2+1=(√2-1)²
x=(√2-1)²/(√2-1)
x=√2-1