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

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Answer 1 · 2018-03-10T17:34:17

See a solution process below:

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

#y^2 - 5y + color(red)(4) = -4 + color(red)(4)#

#y^2 - 5y + 4 = 0#

Next, factor as:

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

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

Solution 1:

#y - 1 = 0#

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

#y - 0 = 1#

#y = 1#

Solution 2:

#y - 4 = 0#

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

#y - 0 = 4#

#y = 4#

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