Solving a System of Linear Equations: Substitution
1 answer:
You might be interested in
You would expand becausee it isn't equal to anything
(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one
so
for the 3rd power
a=x
b=3
=
(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one
so
for the 3rd power
a=x
b=3
Now, for z to be an integer, the sum x+y must be a divisor of 10.
It has to be a positive divisor since z≥0. Also x≠y.
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.
y can be 0,1,2,3,4 or 5 - 6 solutions
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.
2+2+6+10=20 solutions in total.