Systems of Nonlinear Equations: Problems with Solutions

Solution:
Adding the two equations, we get
[tex]x^2+2xy+y^2=15+10\iff (x+y)^2=25[/tex]
The system does have solutions: it can be written as [tex]x(x+y)=15,\ y(x+y)=10[/tex]; for [tex]x+y=5[/tex] this gives [tex](3, 2)[/tex], and for [tex]x+y=-5[/tex] it gives [tex](-3, -2)[/tex]. For both solutions [tex](x+y)^2=25[/tex].

Solution:
Let the sides of the rectangle be [tex]a[/tex] and [tex]b[/tex] (in feet), [tex]a\gt 0,\ b\gt 0[/tex]. The squares have areas [tex]a^2[/tex], [tex]b^2[/tex] and perimeters [tex]4a[/tex], [tex]4b[/tex], so
[tex]a^2+b^2=116,\quad 4a+4b=56\Rightarrow a+b=14[/tex]
Since [tex](a+b)^2=a^2+b^2+2ab[/tex], we get
[tex]196=116+2ab\Rightarrow ab=40[/tex]
Such a rectangle exists: [tex]a[/tex] and [tex]b[/tex] are the roots of [tex]t^2-14t+40=0[/tex], i.e. [tex]4[/tex] and [tex]10[/tex]. The area of the rectangle is [tex]ab=40[/tex] square feet.

Solution:
The first equation holds exactly when [tex]x=0[/tex] or [tex]x=2y[/tex]. If [tex]x=0[/tex], the second equation becomes
[tex]-y^2-3y=0\iff y(y+3)=0\Rightarrow y=0\text{ or }y=-3[/tex]
This gives the pairs [tex](0, 0)[/tex] and [tex](0, -3)[/tex]. If [tex]x=2y[/tex], then
[tex]4y^2-y^2-3y=0\iff 3y(y-1)=0\Rightarrow y=0\text{ or }y=1[/tex]
This gives the pairs [tex](0, 0)[/tex] and [tex](2, 1)[/tex]. The pair [tex](0, 0)[/tex] appears in both cases, so it is counted only once. The solutions are [tex](0, 0)[/tex], [tex](0, -3)[/tex] and [tex](2, 1)[/tex], i.e. the system has [tex]3[/tex] solutions.

Solution:
Multiply the first equation by [tex]3[/tex] and add it to the second one; the terms with [tex]y^2[/tex] and [tex]y[/tex] cancel:
[tex]3x^2+3y^2-15x-3y+2x^2-3y^2+x+3y=3\iff 5x^2-14x-3=0[/tex]
[tex]x_{1,2}=\frac{14\pm 16}{10}\Rightarrow x=3\text{ or }x=-\frac{1}{5}[/tex]
The system is equivalent to this equation together with the first equation. For [tex]x=3[/tex] the first equation gives
[tex]9+y^2-15-y=0\iff y^2-y-6=0\Rightarrow y=3\text{ or }y=-2[/tex]
For [tex]x=-\frac{1}{5}[/tex] it gives [tex]\frac{1}{25}+y^2+1-y=0[/tex], i.e. [tex]y^2-y+\frac{26}{25}=0[/tex], whose discriminant is [tex]1-\frac{104}{25}\lt 0[/tex], so there is no real [tex]y[/tex]. The solutions are [tex](3, 3)[/tex] and [tex](3, -2)[/tex].

Solution:
Factor both equations. Since [tex]x^2+x-12=(x-3)(x+4)[/tex] and [tex]xy-3y=y(x-3)[/tex], the first equation becomes
[tex]y(x-3)-(x-3)(x+4)=0\iff (x-3)(y-x-4)=0[/tex]
In the second equation [tex]y^2+y-2=(y-1)(y+2)[/tex] and [tex]-xy+x=-x(y-1)[/tex], so it becomes
[tex](y-1)(y+2-x)=0[/tex]
Each equation splits into two cases, which gives four simple systems. If [tex]x=3,\ y=1[/tex], we get [tex](3, 1)[/tex]; if [tex]x=3,\ y=x-2[/tex], we get [tex](3, 1)[/tex] again.
If [tex]y=x+4,\ y=1[/tex], then [tex]x=-3[/tex] and we get [tex](-3, 1)[/tex]; if [tex]y=x+4,\ y=x-2[/tex], then [tex]x+4=x-2[/tex], i.e. [tex]4=-2[/tex], which is impossible. The solutions are [tex](3, 1)[/tex] and [tex](-3, 1)[/tex].

Solution:
Domain: [tex]x\ne 0,\ y\ne 0[/tex]. Bring the first equation to a common denominator and use the second equation:
[tex]\frac{x^2-y^2}{xy}=\frac{5}{6}\Rightarrow \frac{5}{xy}=\frac{5}{6}\Rightarrow xy=6[/tex]
Substitute [tex]y=\frac{6}{x}[/tex] into the second equation:
[tex]x^2-\frac{36}{x^2}=5\iff x^4-5x^2-36=0\iff (x^2-9)(x^2+4)=0[/tex]
Since [tex]x^2+4\gt 0[/tex], we get [tex]x^2=9[/tex], [tex]x=\pm 3[/tex], and [tex]y=\frac{6}{x}[/tex] has the same sign as [tex]x[/tex]: the pairs [tex](3, 2)[/tex] and [tex](-3, -2)[/tex]. Both are in the domain and satisfy the system: [tex]\frac{3}{2}-\frac{2}{3}=\frac{5}{6}[/tex] and [tex]9-4=5[/tex].
The pairs [tex](3, -2)[/tex] and [tex](-3, 2)[/tex] are not solutions: for them [tex]xy=-6[/tex] and the left side of the first equation equals [tex]-\frac{5}{6}[/tex]. The solutions are [tex](3, 2)[/tex] and [tex](-3, -2)[/tex].

Solution:
Let the circumferences of the front and the rear wheel be [tex]x[/tex] in and [tex]y[/tex] in, [tex]x\gt 0,\ y\gt 40[/tex]. The distance is [tex]400[/tex] ft, i.e. [tex]4800[/tex] in, so the wheels make [tex]\frac{4800}{x}[/tex] and [tex]\frac{4800}{y}[/tex] turns. The conditions give the system
[tex]\left\{\begin{array}{l} \frac{4800}{x}-\frac{4800}{y}=10 \\ \frac{4800}{x+40}=\frac{4800}{y-40} \end{array}\right.[/tex]
From the second equation [tex]x+40=y-40[/tex], i.e. [tex]y=x+80[/tex]. Substituting into the first equation:
[tex]\frac{4800}{x}-\frac{4800}{x+80}=10\iff 4800\times 80=10x(x+80)\iff x^2+80x-38400=0[/tex]
[tex]x=-40\pm\sqrt{1600+38400}=-40\pm 200[/tex]
The root [tex]x=-240[/tex] is rejected because [tex]x\gt 0[/tex], so [tex]x=160[/tex] and [tex]y=240[/tex]. Check: [tex]\frac{4800}{160}-\frac{4800}{240}=30-20=10[/tex], and with the changed circumferences both wheels make [tex]\frac{4800}{200}=24[/tex] turns. The circumference of the front wheel is [tex]160[/tex] in.