Hello!
To solve this problem, we will use a system of equations. We will have one number be x and the other y. We will use substitutions to solve for each variable.
x+y=9
x=2y-9
To solve for the two numbers, we need to solve the top equation. The second equation shows that x=2y-9. In the first equation, we can replace 2y-9 for x and solve.
2y-9+y=9
3y-9=9
3y=18
y=6
We now know the value of y. Now we need to find x. We can plug in 6 for y in the second equation to find x.
x=2·6-9
x=12-9
x=3
Just to check, we will plug these two numbers into the first equation.
3+6=9
9=9
Our two numbers are three and six.
I hope this helps!
If s is the side of the square base, the area of the square base is s^2.
The volume of the square base is,
V = (s²) (h)
s² = V/h
s² = 3n³ + 13n² + 16n + 4 / <span>3n + 1
You can do this division by factoring, synthetic division, or by plain division.
Factoring out 3n + 1 from the numerator gives you:
</span>s² = (3n + 1)(n² + 4n + 4) / 3n+1
s² = n² + 4n + 4
Therefore, the area of the square base is <span>n² + 4n + 4.</span>
X can always be used a 1 .
- 5x - 4x + 24 = - 3
9x = 27
x = 3
