Answer:
y-3
Problem:
What is the remainder when the dividend is xy-3, the divisor is y, and the quotient is x-1. ?
Step-by-step explanation:
Dividend=quotient×divisor+remainder
So we have
xy-3=(x-1)×(y)+remainder
xy-3=(xy-y)+remainder *distributive property
Now we just need to figure out what polynomial goes in for the remainder so this will be a true identity.
We need to get rid of minus y so we need plus y in the remainder.
We also need minus 3 in the remainder.
So the remainder is y-3.
Let's try it out:
xy-3=(xy-y)+remainder
xy-3=(xy-y)+(y-3)
xy-3=xy-3 is what we wanted so we are done here.
Step-by-step explanation:
The slope of the line is -4/7.
Answer:
23) x= 3, y = 4
24) Any (x,y) values where x = 1.5 - 2y
Step-by-step explanation:
For the first system(23), it goes like this
[1 2 | 7]
[2 1 | 8]
We need to do L2 = L2 - 2L1, so now it is
[1 2 | 7]
[0 -3 |-6]
So, now we have:
-3y = -6 *(-1)
3y = 6
y = 2
x + 2y = 7
x + 4 = 7
x = 3
Now for system 24, we have:
[-1 2 | 1.5]
[2 -4| 3]
We do L2 = L2 + 2L1, so we have:
[-1 2 | 1.5]
[0 0| 0]
So there are infinite solutions for this system. The solution for this system will be each (x,y) pair where x = 1.5 - 2y.
When you multiply it is just 5*23=115
115*2= 230
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