solve for y 3(y+1)=9y-39?

3(y + 1) = 9y - 39

multiply to clear the parentheses
3y + 3 = 9y - 39

subtract 3y from both sides of the equation
3 = 6y - 39

add 39 to both sides of the equation
42 = 6y

divide by 6 on both sides of the equation
7 = y <====ANSWER

Have a great day!

3y+3=9y-39
-6y=-39--3
-6y=-42
y=7

it rather is homogeneous so that's uncomplicated to unravel: wager: y = exp(rx) --> plug in y, y', y'', and y''' into your differential equation r³ + r² + 3r - 5 = 0 --> ought to ingredient...needless to say r = a million is a root: a million + a million + 3 - 5 = 0 <-- r = a million is a root, ingredient this out (do long branch): (r - a million)(ar² + br + c) = r³ + r² + 3r - 5 --> ar³ + br² - ar² + cr - br - c = ar³ + (b - a)r² + (c - b)r - c --> so we've: a = a million b - a= a million --> b = a million + a = 2 c - b = 3 --> c = 3 + b = 5 -c = -5 --> c = +5, examine So we've: r³ + r² + 3r - 5 = (r - a million)(r² + 2r + 5) --> this quadratic has no authentic roots, so the final 2 roots provide imaginary exponential r = -a million ± ?(4 - 20) / 2 = -a million ± 4i/2 = -a million ± 2i --> this supplies a known answer of: y = A * exp(x) + B * exp(-x) * (sin(2x) + C * cos(2x)) @Ted S: rather what I wrote is right--your way is likewise applicable, btw.

3(y + 1) = 9y - 39
y + 1 = (9y - 39)/3
y + 1 = 9y/3 - 39/3
y + 1 = 3y - 13
y - 3y + 1 = -13
-2y = -13 - 1
y = -14/(-2)
y = 7

Y = 7

Expand to give, 3y + 3 = 9y - 39

then, 39 + 3 = 9y - 3y

so, 42 = 6y...........giving y = 7

3y+3=9y-39
3+39=9y-3y
42=6y
y=7

3(y+1)=9y-39

==> 3y + 3 = 9y - 39
==> 9y -3y = 3+ 39
==> 6y = 42
==> y = 7

Y=7

Edit: I'll show the work.

3(y+1)=9y-39
3y + 3=9y-39
3y+42=9y
42=6y
7=y

3y + 3 = 9y - 39

3 + 39 = 9y - 3y

42 = 6y

y = 42/6

y = 7