What is the answer to n(n-4)+n(n+8)=n(n-13)+n(n+1)+16 ?

The answer to n(n-4)+n(n+8)=n(n-13)+n(n+1)+16 is n=1

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n(n-4)+n(n+8)=n(n-13)+n(n+1)+16

n (n - 4) + n (n + 8) = n (n - 13) + n (n + 1) + 16

n = 1

Solution steps

n (n - 4) + n (n + 8) = n (n - 13) + n (n + 1) + 16

Expand

n (n - 4) + n (n + 8) : 2 n^{2} + 4 n

Expand

n (n - 13) + n (n + 1) + 16 : 2 n^{2} - 12 n + 16

2 n^{2} + 4 n = 2 n^{2} - 12 n + 16

Subtract

2 n^{2} - 12 n

from both sides

2 n^{2} + 4 n - (2 n^{2} - 12 n) = 2 n^{2} - 12 n + 16 - (2 n^{2} - 12 n)

Simplify

16 n = 16

Divide both sides by

16

n = 1

\frac{( 2 x + 1 )}{x - 1} > 1

\frac{35}{28} = \frac{n}{12}

\frac{1}{4} - \frac{7}{8}

- \frac{3}{4} x \geq 30

4^{\frac{1}{3}} \cdot 4^{\frac{1}{5}}

What is the answer to n(n-4)+n(n+8)=n(n-13)+n(n+1)+16 ?
The answer to n(n-4)+n(n+8)=n(n-13)+n(n+1)+16 is n=1
Solve for n: n(n-4)+n(n+8)=n(n-13)+n(n+1)+16
The solution is n=1