What is the answer to solvefor x,y[n]-1/2*y[n-1]=x[n] ?

The answer to solvefor x,y[n]-1/2*y[n-1]=x[n] is x=(ny+y)/(2n);\quad n\ne 0

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solvefor x,y[n]-1/2*y[n-1]=x[n]

solve for

x, y [n] - \frac{1}{2} \cdot y [n - 1] = x [n]

x = \frac{n y + y}{2 n}; n \neq = 0

Solution steps

y [n] - \frac{1}{2} y [n - 1] = x [n]

Switch sides

x [n] = y [n] - \frac{1}{2} y [n - 1]

x [n] = n y - \frac{1}{2} y (n - 1)

Divide both sides by

n; n \neq = 0

x = \frac{n y + y}{2 n}; n \neq = 0

9 + x = b

\frac{( 2 + x + 31 )}{7} = 24

\frac{180}{54} = \frac{( 5 x )}{3}

2^{3.5} - (x + 3) = 10

- 4 x = - 100

What is the answer to solvefor x,y[n]-1/2*y[n-1]=x[n] ?
The answer to solvefor x,y[n]-1/2*y[n-1]=x[n] is x=(ny+y)/(2n);\quad n\ne 0