How do you solve y² = x + 3 and x - 2y = 12?

3 Answers
Jun 10, 2018

Answer:

If #y=5#, #x=22#
If #y=-3#, #x=6#

Explanation:

#y^2=x+3# --- (1)
#x-2y=12# --- (2)

From (2)
#x-2y=12#
#x=12+2y# --- (3)

Sub (3) into (1)
#y^2=12+2y+3#
#y^2=15+2y#
#y^2-2y-15=0#
#(y-5)(y+3)=0#
#y=5# or #y=-3#

If #y=5#,
#x=12+2y#
#x=12+10#
#x=22#

If #y=-3#,
#x=12+2y#
#x=12-6#
#x=6#

Jun 10, 2018

Answer:

#(6,-3),(22,5)#

Explanation:

#y^2=x+3to(1)#

#x-2y=12to(2)#

#"from equation "(2)tox=12+2yto(3)#

#"substitute "x=12+2y" in equation "(1)#

#y^2=12+2y+3#

#y^2-2y-15=0larrcolor(blue)"in standard form"#

#(y-5)(y+3)=0#

#y+3=0rArry=-3#

#y-5=0rArry=5#

#"substitute these values into equation "(3)#

#y=-3tox=12-6=6to(6,-3)#

#y=5tox=12+10=22to(22,5)#

Answer:

A graphical solution below:

Explanation:

Solving this graphically:

Here's one point of intersection:

graph{(y^2-x-3)(x-2y-12)=0}

And here's the other:

graph{(y^2-x-3)(x-2y-12)=0[10,30,0,10]}