How do you solve the simultaneous equations #x^2 + y^2=29# and #y-x=3#?

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28
Jul 19, 2015

Answer:

Use the second equation to provide an expression for #y# in terms of #x# to substitute into the first equation to give a quadratic equation in #x#.

Explanation:

First add #x# to both sides of the second equation to get:

#y = x+3#

Then substitute this expression for #y# into the first equation to get:

#29 = x^2+(x+3)^2 = 2x^2+6x+9#

Subtract #29# from both ends to get:

#0 = 2x^2+6x-20#

Divide both sides by #2# to get:

#0 = x^2+3x-10 = (x+5)(x-2)#

So #x=2# or #x=-5#

If #x=2# then #y = x+3 = 5#.

If #x=-5# then #y = x+3 = -2#

So the two solutions #(x, y)# are #(2, 5)# and #(-5, -2)#

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Sep 10, 2016

Answer:

#(x=-5 and y=-2) or (x=2 and y=5)#

Explanation:

Since you have both #x^2+y^2=29# and #y-x=3#,

You want to combine these two equations into one equation with a single variable, solve it and then solve for the other variable. An example on how to do this goes like this:

#y-x=3 rarr y=x+3 # and we have # y^2=x^2+6x+9#

Since #x^2+y^2=29#, substitute the expression for #y^2# into this:

#2x^2+6x+9=29#, so #2x^2+6x-20=0#.

We can solve for #x# using the quadratic formula:
#x=(-6pmsqrt(36-4*2*(-20)))/(2*2)=-3/4pm1/4sqrt(196)=(-6pm14)/4#
So #x=-5# or #x=2#.

Since #y=x+3#, this gives #(x=-5 and y=-2) or (x=2 and y=5)#.

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