What is the answer to solvefor y,r(t)=(t^2-2t)y+(3t+6)j ?

The answer to solvefor y,r(t)=(t^2-2t)y+(3t+6)j is y=(rt-j(3t+6))/(t^2-2t);\quad t\ne 2,t\ne 0

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solvefor y,r(t)=(t^2-2t)y+(3t+6)j

solve for

y, r (t) = (t^{2} - 2 t) y + (3 t + 6) j

y = \frac{r t - j ( 3 t + 6 )}{t ^{2} - 2 t}; t \neq = 2, t \neq = 0

Solution steps

r (t) = (t^{2} - 2 t) y + (3 t + 6) j

Switch sides

(t^{2} - 2 t) y + (3 t + 6) j = r t

(t^{2} - 2 t) y + (3 t + 6) j = r t

Move

(3 t + 6) j

to the right side

(t^{2} - 2 t) y = r t - (3 t + 6) j

Divide both sides by

t^{2} - 2 t; t \neq = 2, t \neq = 0

y = \frac{r t - j ( 3 t + 6 )}{t ^{2} - 2 t}; t \neq = 2, t \neq = 0

z, \frac{p}{10} + \frac{z}{3} = 100

a (n) = 450 + (n - 1) (0.05 \cdot 450), 765

h = \frac{1}{9} + \frac{1}{4}

y - 6 y + 13 y = 0

y + 5 = 17, y

What is the answer to solvefor y,r(t)=(t^2-2t)y+(3t+6)j ?
The answer to solvefor y,r(t)=(t^2-2t)y+(3t+6)j is y=(rt-j(3t+6))/(t^2-2t);\quad t\ne 2,t\ne 0