the sum of the digits of a two digit number is 11.the number obtained by interchanging the digits exceeds the given number by 45.find the number
jeniselva13:
how we get -9u-9t=-99?
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t = 10's digit
u = 1's digit
t + u = 11
10u + t = 45 + (10t + u)
Combine like terms: 9u - 9t = 45
Solve by elimination:
9u - 9t = 45
Now, we know that u +t = 11
multiply -9 on both sides...
-9(u+t)=-99
-9u -9t = -99
-18t = -54
t = 3, u = 8
Original number is 38.
u = 1's digit
t + u = 11
10u + t = 45 + (10t + u)
Combine like terms: 9u - 9t = 45
Solve by elimination:
9u - 9t = 45
Now, we know that u +t = 11
multiply -9 on both sides...
-9(u+t)=-99
-9u -9t = -99
-18t = -54
t = 3, u = 8
Original number is 38.
can u expln how we get -9u-9t=-99?
By multiplying the u+t=11 equation by -9 on both sides
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