Systems of Nonlinear Equations: Difficult Problems with Solutions

Solution:
The first equation must not be divided by [tex]x-y[/tex], because this expression may be zero and solutions would be lost. Move everything to the left side and factor:
[tex](x-y)(x^2+xy+y^2)-8(x-y)=0\iff (x-y)(x^2+xy+y^2-8)=0[/tex]
If [tex]x=y[/tex], the second equation gives [tex]2x^2=8[/tex], so [tex]x=\pm 2[/tex]: the pairs [tex](2, 2)[/tex] and [tex](-2, -2)[/tex].
If [tex]x^2+xy+y^2=8[/tex], subtracting the second equation [tex]x^2+y^2=8[/tex] gives [tex]xy=0[/tex]. For [tex]x=0[/tex] we get [tex]y^2=8[/tex], [tex]y=\pm 2\sqrt{2}[/tex]; for [tex]y=0[/tex] we get [tex]x=\pm 2\sqrt{2}[/tex].
All six pairs satisfy both equations, so the solutions are
[tex](2, 2),\ (-2, -2),\ (0, 2\sqrt{2}),\ (0, -2\sqrt{2}),\ (2\sqrt{2}, 0),\ (-2\sqrt{2}, 0)[/tex]

Solution:
Add the two equations and put [tex]u=x+y[/tex]:
[tex]x^2+2xy+y^2+(x+y)=72\iff u^2+u-72=0\Rightarrow u=8\text{ or }u=-9[/tex]
If [tex]x+y=-9[/tex], the second equation gives [tex]xy=\frac{38+9}{2}=\frac{47}{2}[/tex]. Then [tex]x[/tex] and [tex]y[/tex] would be roots of [tex]t^2+9t+\frac{47}{2}=0[/tex], whose discriminant is [tex]81-94\lt 0[/tex], so this case gives no real solutions.
If [tex]x+y=8[/tex], then [tex]xy=\frac{38-8}{2}=15[/tex], and by Vieta's formulas [tex]x[/tex] and [tex]y[/tex] are the roots of
[tex]t^2-8t+15=0\Rightarrow t_1=3,\ t_2=5[/tex]
The solutions are [tex](3, 5)[/tex] and [tex](5, 3)[/tex], so [tex]\frac{x}{y}[/tex] equals [tex]\frac{3}{5}[/tex] or [tex]\frac{5}{3}[/tex]. The largest value is [tex]\frac{5}{3}[/tex].

Solution:
Factor the left side of the first equation:
[tex]x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=(x^2+xy+y^2)(x^2-xy+y^2)[/tex]
Using the second equation, the first one becomes [tex]37(x^2-xy+y^2)=481[/tex], so [tex]x^2-xy+y^2=13[/tex]. Hence the system is equivalent to
[tex]\left\{\begin{array}{l} x^2+xy+y^2=37 \\ x^2-xy+y^2=13 \end{array}\right.[/tex]
Subtracting the equations gives [tex]2xy=24[/tex], i.e. [tex]xy=12[/tex], and adding them gives [tex]2(x^2+y^2)=50[/tex], i.e. [tex]x^2+y^2=25[/tex]. Then
[tex](x+y)^2=x^2+y^2+2xy=25+24=49\Rightarrow x+y=\pm 7[/tex]
If [tex]x+y=7[/tex] and [tex]xy=12[/tex], then [tex]x[/tex] and [tex]y[/tex] are the roots of [tex]t^2-7t+12=0[/tex], i.e. [tex]3[/tex] and [tex]4[/tex].
If [tex]x+y=-7[/tex] and [tex]xy=12[/tex], they are the roots of [tex]t^2+7t+12=0[/tex], i.e. [tex]-3[/tex] and [tex]-4[/tex].
The solutions are [tex](3, 4)[/tex], [tex](4, 3)[/tex], [tex](-3, -4)[/tex], [tex](-4, -3)[/tex].

Solution:
The second equation is [tex]xy(x-y)=6[/tex]. We use the identity
[tex]x^{3}-y^{3}=(x-y)^{3}+3xy(x-y)[/tex]
Then the first equation becomes
[tex](x-y)^{3}+3\times 6=26\Rightarrow (x-y)^{3}=8[/tex]
The real cube root is unique, so [tex]x-y=2[/tex] and then [tex]xy=\frac{6}{2}=3[/tex]. With [tex]x=y+2[/tex] we get
[tex]y(y+2)=3\Rightarrow y^{2}+2y-3=0\Rightarrow y_{1}=1,\ y_{2}=-3[/tex]
so [tex]x_{1}=3[/tex], [tex]x_{2}=-1[/tex]. Check: [tex]27-1=26[/tex], [tex]3\times 2=6[/tex] and [tex]-1+27=26[/tex], [tex]3\times 2=6[/tex].
The solutions are [tex](3, 1)[/tex] and [tex](-1, -3)[/tex].

Solution:
Since [tex]5\times 44=4\times 55=220[/tex], multiply the first equation by [tex]5[/tex], the second by [tex]4[/tex] and subtract:
[tex]10x^2-5xy+25y^2-(4x^2-12xy+28y^2)=0\Rightarrow 6x^2+7xy-3y^2=0\Rightarrow (3x-y)(2x+3y)=0[/tex]
This equation together with the first equation is equivalent to the system. Hence [tex]y=3x[/tex] or [tex]x=-\frac{3y}{2}[/tex].
If [tex]y=3x[/tex], the first equation gives [tex]2x^2-3x^2+45x^2=44x^2=44[/tex], so [tex]x=\pm 1[/tex]: the pairs [tex](1, 3)[/tex] and [tex](-1, -3)[/tex].
If [tex]x=-\frac{3y}{2}[/tex], the first equation gives [tex]\frac{9y^2}{2}+\frac{3y^2}{2}+5y^2=11y^2=44[/tex], so [tex]y=\pm 2[/tex]: the pairs [tex](-3, 2)[/tex] and [tex](3, -2)[/tex].
Check in the second equation: [tex]1-9+63=55[/tex] for [tex](1, 3)[/tex] and [tex]9+18+28=55[/tex] for [tex](-3, 2)[/tex]; the opposite pairs give the same values.
The solutions are [tex](1, 3)[/tex], [tex](-1, -3)[/tex], [tex](3, -2)[/tex], [tex](-3, 2)[/tex].

Solution:
Multiply the first equation by [tex]2[/tex] and factor:
[tex]2x^2-5xy+2y^2=0\Rightarrow (x-2y)(2x-y)=0[/tex]
Hence [tex]x=2y[/tex] or [tex]y=2x[/tex].
If [tex]x=2y[/tex], the second equation becomes [tex]y=\frac{2y^2}{4}[/tex], i.e. [tex]y(y-2)=0[/tex], so [tex]y=0[/tex] or [tex]y=2[/tex]: the pairs [tex](0, 0)[/tex] and [tex](4, 2)[/tex].
If [tex]y=2x[/tex], the second equation becomes [tex]-x=\frac{2x^2}{4}[/tex], i.e. [tex]x(x+2)=0[/tex], so [tex]x=0[/tex] or [tex]x=-2[/tex]: the pairs [tex](0, 0)[/tex] and [tex](-2, -4)[/tex].
Check: for [tex](4, 2)[/tex], [tex]16+4=20=\frac{5}{2}\times 8[/tex] and [tex]4-2=2=\frac{8}{4}[/tex]; for [tex](-2, -4)[/tex], [tex]4+16=20=\frac{5}{2}\times 8[/tex] and [tex]-2+4=2=\frac{8}{4}[/tex]; [tex](0, 0)[/tex] obviously works.
The solution [tex](0, 0)[/tex] is easily lost if one divides by [tex]xy[/tex] or by [tex]x-y[/tex], which can be [tex]0[/tex]. The solutions are [tex](0, 0)[/tex], [tex](4, 2)[/tex], [tex](-2, -4)[/tex].

Solution:
A square root is non-negative, so [tex]x\ge 0[/tex] and [tex]y\ge 0[/tex]. Under these conditions the system is equivalent to
[tex]\left\{\begin{array}{l} x^2+5x-6=y^2 \\ y^2+5y-6=x^2 \end{array}\right.[/tex]
Subtracting the second equation from the first:
[tex]x^2-y^2+5(x-y)=y^2-x^2\Rightarrow 2(x-y)(x+y)+5(x-y)=0\Rightarrow (x-y)(2x+2y+5)=0[/tex]
Since [tex]x,y\ge 0[/tex], we have [tex]2x+2y+5\gt 0[/tex], hence [tex]x=y[/tex]. Then [tex]x^2+5x-6=x^2[/tex], so [tex]x=\frac{6}{5}[/tex].
Check: [tex]x^2+5x-6=\frac{36}{25}[/tex] and [tex]\sqrt{\frac{36}{25}}=\frac{6}{5}=y[/tex], so [tex]x=y=\frac{6}{5}[/tex] satisfies the system. Therefore
[tex]xy=\frac{36}{25}[/tex]

Solution:
Domain: [tex]x\ne 0[/tex], [tex]y\ne 0[/tex], [tex]x-y\gt 0[/tex], [tex]x+y\gt 0[/tex]. Since [tex]4^{t}=2^{2t}[/tex] and [tex]32=2^5[/tex], the first equation gives
[tex]2\left(\frac{x}{y}+\frac{y}{x}\right)=5[/tex]
Put [tex]t=\frac{x}{y}[/tex]:
[tex]t+\frac{1}{t}=\frac{5}{2}\Rightarrow 2t^2-5t+2=0[/tex]
so [tex]t=2[/tex] or [tex]t=\frac{1}{2}[/tex], i.e. [tex]x=2y[/tex] or [tex]y=2x[/tex]. The second equation is [tex]\log_3(x-y)+\log_3(x+y)=1[/tex], i.e. [tex](x-y)(x+y)=3[/tex], so
[tex]x^2-y^2=3[/tex]
If [tex]y=2x[/tex], then [tex]x^2-4x^2=-3x^2=3[/tex], which is impossible. If [tex]x=2y[/tex], then [tex]3y^2=3[/tex], so [tex]y=\pm 1[/tex].
For [tex]y=1[/tex], [tex]x=2[/tex] we have [tex]x-y=1\gt 0[/tex] and [tex]x+y=3\gt 0[/tex], so the pair is admissible. For [tex]y=-1[/tex], [tex]x=-2[/tex] we have [tex]x-y=-1\lt 0[/tex], so [tex]\log_3(x-y)[/tex] is not defined and the pair is rejected.
Check: [tex]4^{2+\frac{1}{2}}=4^{\frac{5}{2}}=32[/tex] and [tex]\log_3 1=0=1-\log_3 3[/tex]. The solution is [tex](2, 1)[/tex].

Solution:
Write the system as [tex]\sqrt{5y-x}=3-x[/tex], [tex]\sqrt{2y-x}=3-x-y[/tex]. A square root is non-negative, so we need [tex]3-x\ge 0[/tex] and [tex]3-x-y\ge 0[/tex]; under these conditions squaring is an equivalent step:
[tex]\left\{\begin{array}{l} 5y-x=(3-x)^2 \\ 2y-x=(3-x-y)^2 \end{array}\right.[/tex]
Subtracting the second equation from the first:
[tex]3y=(3-x)^2-(3-x-y)^2=y(6-2x-y)[/tex]
If [tex]y=0[/tex], the first equation becomes [tex]-x=(3-x)^2[/tex], i.e. [tex]x^2-5x+9=0[/tex], whose discriminant [tex]25-36[/tex] is negative, so there is no solution.
If [tex]y\ne 0[/tex], then [tex]6-2x-y=3[/tex], i.e. [tex]y=3-2x[/tex]. Substituting into [tex]5y-x=(3-x)^2[/tex]:
[tex]15-11x=9-6x+x^2\Rightarrow x^2+5x-6=0\Rightarrow x_1=1,\ x_2=-6[/tex]
For [tex]x=1[/tex]: [tex]y=1[/tex], [tex]3-x=2\ge 0[/tex], [tex]3-x-y=1\ge 0[/tex], and indeed [tex]\sqrt{4}+1=3[/tex], [tex]\sqrt{1}+1+1=3[/tex].
For [tex]x=-6[/tex]: [tex]y=15[/tex], but [tex]3-x-y=-6\lt 0[/tex]; indeed [tex]\sqrt{2y-x}+x+y=\sqrt{36}+9=15\ne 3[/tex], so this pair is an extraneous root introduced by squaring.
The only solution is [tex](1, 1)[/tex].

Solution:
The right side of the second equation must be non-negative: [tex]x+y\ge 0[/tex]. Under this condition squaring is an equivalent step (then [tex]x+y^{2}=(x+y)^{2}\ge 0[/tex] holds automatically):
[tex]x+y^{2}=x^{2}+2xy+y^{2}\Rightarrow x^{2}+2xy-x=0\Rightarrow x(x+2y-1)=0[/tex]
If [tex]x=0[/tex], the condition becomes [tex]y\ge 0[/tex] and the first equation gives
[tex]2\times 2^{y+1}=9\Rightarrow 2^{y+2}=9\Rightarrow y=\log_{2}9-2=\log_{2}\frac{9}{4}[/tex]
Since [tex]\frac{9}{4}\gt 1[/tex], we have [tex]y\gt 0[/tex], so [tex]\left(0, \log_{2}\frac{9}{4}\right)[/tex] is a solution.
If [tex]x=1-2y[/tex], the condition [tex]x+y=1-y\ge 0[/tex] gives [tex]y\le 1[/tex]. The first equation becomes [tex]2^{1-2y}\times 2^{y+1}+2^{y+1}=9[/tex], i.e. [tex]2^{2-y}+2\times 2^{y}=9[/tex]. Put [tex]t=2^{y}\gt 0[/tex]:
[tex]\frac{4}{t}+2t=9\Rightarrow 2t^{2}-9t+4=0\Rightarrow t_{1}=4,\ t_{2}=\frac{1}{2}[/tex]
For [tex]t=4[/tex] we get [tex]y=2\gt 1[/tex], which is rejected: the pair [tex](-3, 2)[/tex] gives [tex]\sqrt{1}=1\ne -1[/tex] (an extraneous root from squaring).
For [tex]t=\frac{1}{2}[/tex] we get [tex]y=-1[/tex], [tex]x=3[/tex]. Check: [tex](8+1)\times 2^{0}=9[/tex] and [tex]\sqrt{3+1}=2=3+(-1)[/tex].
The solutions are [tex]\left(0, \log_{2}\frac{9}{4}\right)[/tex] and [tex](3, -1)[/tex].

Solution:
Since [tex]a^{2}+b^{2}+c^{2}=1[/tex], each square is at most [tex]1[/tex], so [tex]a\le 1[/tex], [tex]b\le 1[/tex], [tex]c\le 1[/tex]. Subtract the second equality from the first:
[tex]a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c)=0[/tex]
Each term is the product of a square and a non-negative number, so it is non-negative. A sum of non-negative numbers equals [tex]0[/tex] only if every term is [tex]0[/tex], hence [tex]a^{2}(1-a)=0[/tex], i.e. [tex]a=0[/tex] or [tex]a=1[/tex]; the same holds for [tex]b[/tex] and [tex]c[/tex].
For [tex]t=0[/tex] and [tex]t=1[/tex] we have [tex]t^{2}=t[/tex], therefore
[tex]a+b+c=a^{2}+b^{2}+c^{2}=1[/tex]
(In fact, exactly one of the numbers equals [tex]1[/tex] and the other two equal [tex]0[/tex].)