Given: xy+4y=16,x+y=6
⇒y(x+4)=16,x+y=6
⇒(6−x)(x+4)=16
⇒6x+24−x2−4x−16=0
⇒−x2+2x+8=0
⇒x2−2x−8=0
⇒(x−4)(x+2)=0
⇒x=4,−2

So, the required area is given by,
A=∫−24((6−x)−(x+416))dx
⇒A=[6x−2x2−16log∣x+4∣]−24
⇒A=(24−8−16log∣8∣)−(−12−2−16log2)
⇒A=16−16log∣8∣+14+16log2
⇒A=30−32ln2