Odpowiedź:
Zadanie 1.
(I)
[tex]\frac{ex}{sx+a} =t\\ex=t(sx+a)\\ex=tsx+ta\\ex-tsx=ta\\(e-ts)x=ta\\x=\frac{ta}{e-ts} \\x=\frac{at}{e-st}[/tex]
(II)
[tex]\frac{ex}{jx+l} =a\\ex=a(jx+l)\\ex=ajx+al\\ex-ajx=al\\x(e-aj)=al\\x=\frac{al}{e-aj} \\x=\frac{al}{e-ja}[/tex]
(III)
[tex]\frac{tx}{rx+s} =a\\tx=a(rx+s)\\tx=arx+as\\tx-arx=as\\x(t-ar)=as\\x=\frac{as}{t-ar} \\x=\frac{sa}{t-ra}[/tex]
Zadanie 2
(I)
[tex]\frac{ex+t}{kx+p} =a\\ex+t=a(kx+p)\\ex+t=akx+ap\\ex-akx=ap-t\\x(e-ak)=ap-t\\x=\frac{ap-t}{e-ak} \\x=\frac{ap-t}{e-ka}[/tex]
(II)
[tex]\frac{yx+r}{kx+f} =a\\yx+r=a(kx+f)\\yx+r=akx+af\\yx-akx=af-r\\x(y-ak)=af-r\\x=\frac{af-r}{y-ak} \\x=\frac{af-r}{y-ka}[/tex]
(III)
[tex]\frac{rx+u}{tx+k} =a\\rx+u=a(tx+k)\\rx+u=atx+ak\\rx-atx=ak-u\\x(r-at)=ak-u\\x=\frac{ak-u}{r-at}[/tex]
