if d is HCF of 40 and 65,find the value of x and y to satisfy d=40x+65y
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65÷45⇒65=45×1+20
45÷20⇒45=20×2+5
20÷5⇒20=5×4+0 hcf=5
linear combination:
5=45-20×2
5=45-(65-45×1)×2
5=45-65×2+45×1×2
5=45(1+2)-65×2
d=40x+65y
x=3,y=-2
5=45×3-65×2+45×65-45×65
45×3+45×65-65×2-45×65
45(3+65)-65(2+45)
45×69-65×47
x=69,y=47
45÷20⇒45=20×2+5
20÷5⇒20=5×4+0 hcf=5
linear combination:
5=45-20×2
5=45-(65-45×1)×2
5=45-65×2+45×1×2
5=45(1+2)-65×2
d=40x+65y
x=3,y=-2
5=45×3-65×2+45×65-45×65
45×3+45×65-65×2-45×65
45(3+65)-65(2+45)
45×69-65×47
x=69,y=47
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