State a cubic or quartic function with the least degree possible in intercept form for the given graph. Assume that all x-interc
epts are integers, and that the constant factor a is either 1 or −1.
Enter the correct value in the box to answer the question.
y= ?
Enter the correct value in the box to answer the question.
y= ?
1 answer:
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Basically you're solving for both variables here.
I prefer elimination method so that is what I used.
I started off by multiplying each equation in order to get one of the variables at the same value so it would be possible to cancel it out.
Multiplying the first equation by 3 gave me,
12x + 15y = 69
Then I multiplied the second equation by 4,
-12x + 28y = -56
As you can see, it's not possible to cancel out the x variable.
12x + 15y = 69
+(-12x + 28y = -56)
_____________
13y = 13
Then just solve for y which gives you -1.
After you have one variable solved simply insert it into one of the original equations to find the other variable.
4x + 5(-1) = 23
4x - 5 = 23
4x = 28
x = 7.
And there you have it! Hope this helped!
I prefer elimination method so that is what I used.
I started off by multiplying each equation in order to get one of the variables at the same value so it would be possible to cancel it out.
Multiplying the first equation by 3 gave me,
12x + 15y = 69
Then I multiplied the second equation by 4,
-12x + 28y = -56
As you can see, it's not possible to cancel out the x variable.
12x + 15y = 69
+(-12x + 28y = -56)
_____________
13y = 13
Then just solve for y which gives you -1.
After you have one variable solved simply insert it into one of the original equations to find the other variable.
4x + 5(-1) = 23
4x - 5 = 23
4x = 28
x = 7.
And there you have it! Hope this helped!