Question #30c88

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Answer 1 · 2016-10-24T22:47:18

#(x, y) in {(0, 0), (-3, 2)}#

#{(x+2y = 1),(y^2=1-x):}#

In the second equation, add #x-y^2# to both sides to isolate #x#:

#x = 1-y^2#

Substitute this value for #x# into the first equation.

#(1-y^2)+2y = 1#

Gather the terms on the left hand side.

#-y^2+2y+1-1 = 0#

Combine like terms and multiply by #-1# to get a coefficient of #1# for #y^2#.

#y^2-2y = 0#

This can be factored easily, so we will solve by factoring

#y(y-2) = 0#

#=> y = 0 or y-2 = 0#

#=> y = 0 or y = 2#

Substitute the possible solutions for #y# back into #x = 1-y^2# to find the #x# that goes with them.

#y = 0 => x = 1-0^2 = 1#

#y = 2 => x = 1-2^2 = -3#

Thus, our potential solutions are

#(x, y) = (1, 0) or (x, y) = (-3, 2)#

Checking these answers by substituting them into the original equations, we find they both satisfy the system. Thus, our final result is:

#(x, y) in {(1, 0), (-3, 2)}#