In the blank the answer is x axis
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5
II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7
then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15
3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1
insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4
insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2
so the solution is
x=2
y=-4
z=-1
Answer:
8, 10, 3
Step-by-step explanation:
For finding real solutions to cubic equations and those of higher degree, I find a graphing calculator to be useful. The problem can be cast in the form ...
f(x) = 0
where f(x) is the difference between the product of the integers and 240:
f(x) = x(x +2)(x -5) -240
Then all we need to do is ask the calculator to show the x-intercept. It is x=8.
So, the three integers are ...
It is not an equilateral triangle. To prove this, you must solve for x. I found x by setting 2x+5.2 = 3x+1.2 to get x=4. Then I plugged 4 into equation to see if they were equal. 2(4)+5.2= 13.2 and 3(4)+1.3=13.2 but 2(4)+6.2=14.2. This allows us to conclude that the triangle is not equilateral because all of the sides are NOT congruent. Hope that helped:))
