First equation is x-3y+2z=11
We coundt x and we've got
x=11+3y-2z
Third equation is 2x-2y-4z=2
We count also x and we've got:
2x=2+2y+4z /:2
x=1+y+2z
so:
11+3y-2z=1+y+2z
2y-4z=-10 /:2
y-2z=-5
y=2z-5so
x=1+y+2z
x=1+2z-5+2z
x=4z-4
Now substitute this to second equation:
-x+4y+3z=5
-(4z-4) + 4(2z-5) +3z=5
-4z+4+8z-20+3z=5
7z-16=5
7z=21 /:7
z=3
And count rest:
x=4z-4 = 4*3-4=8
y=2z-5=2*3-5=1
So the solution is:
Answer:
I believe the answer is 4
Step-by-step explanation:
Answer: y=5x-5
Step-by-step explanation:
<span>The earnings by a stock invested at r% for n years is obtained by the formular A = P(1 + r)^n; where P is the initial investment = 1,500; r is the interest rate = 10% and n is the numberof years of the investment. Here A = 1,500(1 + 0.1)^18 = 1,500(1.1)^18 = 8,339.88.Hope this helps. Let me know if you need additional help!</span>