How do you solve $y^{2} - 5 y = - 4$ $y^2 - 5y = -4$ ?

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Answer 1 · 2018-03-10T17:34:17

See a solution process below:

First, add $4$ $color(red)(4)$ to each side of the equation to put it into standard quadratic form:

$y^{2} - 5 y + 4 = - 4 + 4$ $y^2 - 5y + color(red)(4) = -4 + color(red)(4)$

$y^{2} - 5 y + 4 = 0$ $y^2 - 5y + 4 = 0$

Next, factor as:

$(y - 1) (y - 4) = 0$ $(y - 1)(y - 4) = 0$

Now, solve each term on the left for $0$ $0$ :

Solution 1:

$y - 1 = 0$ $y - 1 = 0$

$y - 1 + 1 = 0 + 1$ $y - 1 + color(red)(1) = 0 + color(red)(1)$

$y - 0 = 1$ $y - 0 = 1$

$y = 1$ $y = 1$

Solution 2:

$y - 4 = 0$ $y - 4 = 0$

$y - 4 + 4 = 0 + 4$ $y - 4 + color(red)(4) = 0 + color(red)(4)$

$y - 0 = 4$ $y - 0 = 4$

$y = 4$ $y = 4$

The Solution Set Is: $y = {1, 4}$ $y = {1, 4}$

How do you solve y^2 - 5y = -4y2−5y=−4y^2 - 5y = -4?