3. True or False, the following 4 ordered pairs are solutions to the following equation:
1 answer:
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Answer:
Quotient: 5x+11
Remainder: 59
Step-by-step explanation:
I'm going to do long division.
The bottom goes on the outside and the top goes in the inside.
Setup:
---------------------------------
2x-5 | 10x^2 -3x +4
Starting the problem from the setup:
5x +11 (I put 5x on top because 5x(2x)=10x^2)
--------------------------------- (We are going to distribute 5x to the divisor)
2x-5 | 10x^2 -3x +4
-(10x^2 -25x) (We are now going to subtract to see what's left.)
-----------------------------------
22x +4 (I know 2x goes into 22x, 11 times.)
( I have put +11 on top as a result.)
-(22x -55) (I distribute 11 to the divisor.)
-----------------------
59 (We are done since the divisor is higher degree.)
The quotient is 5x+11.
The remainder is 59.
The result of the division is equal to:
.
We can actually use synthetic division as well since the denominator is linear.
Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:
2x-5=0
Add 5 on both:
2x=5
Divide both sides by 2:
x=5/2
Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:
So 5/2 will go on the outside:
5/2 | 10 -3 4
| 25 55
------------------------------
10 22 59
So we have:
Distribute the 1/2 back:
y = 4
first note that the right side simplifies to 12, equation can be written
= 12 ( multiply both sides by 0.8 )
2y + 1.6 = 9.6 ( subtract 1.6 from both sides )
2y = 8 ( divide both sides by 2 )
y = 4