How do you find the solution set for 2y=5x+10 and y^2-4y=5x+10?

1 Answer
Aug 11, 2015

Answer:

#{(x = -2), (y=0):}" "# or #" "{(x=2/5), (y=6):}#

Explanation:

Your system of equations looks like this

#{(2y = 5x + 10), (y^2 - 4y = 5x+10):}#

Right from the start, you can say that this system of equations can be written as one equation in #y#, since you have

#(5x+10) = 2y" "# and #" "5x+10 = y^2 - 4y#

This is equivalent to

#underbrace(y^2 - 4y)_(color(blue)(=5x+10)) = overbrace(2y)^(color(orange)(=5x+10))#

Rearrange this equation into classic quadratic form

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

#y^2 -6y = 0#

You can factor this equation to get

#y * (y - 6) = 0#

The two solutions will thus be #y=0# and #y=6#.

Use these values of #y# in one of the two original equations to get the values of #x#.

  • when #y=o#, you have

#5x+10 = 0^2 - 4 * 0#

#5x+10 = 0#

#5x = -10 implies x= (-10)/5 = -2#

  • when #y=6#, you have

#5x+10 = 6^2 - 4 * 6#

#5x + 10 = 12#

#5x = 2 implies x = 2/5#

The two solution sets for this system of equations are

#{(x = -2), (y=0):}" "# or #" "{(x=2/5), (y=6):}#