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Solve each system of nonlinear equations.

\(\textbf{1)}\) \(x^2+y^2=8\)
\(\,\,\,\,\,\,\,\) \(y=x\)

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The solutions are \( (-2,-2), \enspace (2,2) \)

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\(\,\,\,\,\,x^2+y^2=8 \text{ and }y=x\)

\(\,\,\,\,\,x^2+(x)^2=8\)

\(\,\,\,\,\,2x^2=8\)

\(\,\,\,\,\,x^2=4\)

\(\,\,\,\,\,x=\pm\sqrt{4}\)

\(\,\,\,\,\,x=\pm 2\)

\(\,\,\,\,\,y=x\)

\(\,\,\,\,\,(-2,-2), \, (2,2)\)

\(\textbf{2)}\) \(x^2+y^2=25\)
\(\,\,\,\,\,\,\,\) \(y=\frac{3}{4}x\)

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The solutions are \( (-4,-3), \enspace (4,3) \)

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\(\,\,\,\,\,x^2+y^2=25 \text{ and }y=\frac{3}{4}x\)

\(\,\,\,\,\,x^2+\left(\frac{3}{4}x\right)^2=25\)

\(\,\,\,\,\,x^2+\frac{9}{16}x^2=25\)

\(\,\,\,\,\,\frac{25}{16}x^2=25\)

\(\,\,\,\,\,x^2=16\)

\(\,\,\,\,\,x=\pm 4\)

\(\,\,\,\,\,y=\frac{3}{4}x\)

\(\,\,\,\,\,(-4,-3), \,(4,3)\)

\(\textbf{3)}\) \(y=x^2+2\)
\(\,\,\,\,\,\,\,\) \(y=2\)

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The solution is \( (0,2) \)

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\(\,\,\,\,\,x^2+2=2\)

\(\,\,\,\,\,x^2=0\)

\(\,\,\,\,\,x=0\)

\(\,\,\,\,\,y=2\)

\(\,\,\,\,\,(0,2)\)

\(\textbf{4)}\) \(y=x^2-2\)
\(\,\,\,\,\,\,\,\) \(y=2\)

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The solutions are \( (-2,2), \enspace (2,2) \)

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\(\,\,\,\,\,x^2-2=2\)

\(\,\,\,\,\,x^2=4\)

\(\,\,\,\,\,x=\pm 2\)

\(\,\,\,\,\,y=2\)

\(\,\,\,\,\,(-2,2), \,(2,2)\)

\(\textbf{5)}\) \(y=x^2-3\)
\(\,\,\,\,\,\,\,\) \(y=-x^2+5\)

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The solutions are \( (-2,1), \enspace (2,1) \)

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\(\,\,\,\,\,x^2-3=-x^2+5\)

\(\,\,\,\,\,2x^2=8\)

\(\,\,\,\,\,x^2=4\)

\(\,\,\,\,\,x=\pm 2\)

\(\,\,\,\,\,y=x^2-3\)

\(\,\,\,\,\,(-2,1), \,(2,1)\)

\(\textbf{6)}\) \(y=x^2+2\)
\(\,\,\,\,\,\,\,\) \(y=2x+2\)

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The solutions are \( (0,2), \enspace (2,6) \)

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\(\,\,\,\,\,x^2+2=2x+2\)

\(\,\,\,\,\,x^2-2x=0\)

\(\,\,\,\,\,x(x-2)=0\)

\(\,\,\,\,\,x=0,2\)

\(\,\,\,\,\,y=x^2+2\)

\(\,\,\,\,\,(0,2), \,(2,6)\)

\(\textbf{7)}\) \(y=x^2+2\)
\(\,\,\,\,\,\,\,\) \(y=2x+1\)

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The solution is \( (1,3) \)

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\(\,\,\,\,\,x^2+2=2x+1\)

\(\,\,\,\,\,x^2-2x+1=0\)

\(\,\,\,\,\,(x-1)^2=0\)

\(\,\,\,\,\,x=1\)

\(\,\,\,\,\,y=3\)

\(\,\,\,\,\,(1,3)\)

\(\textbf{8)}\) \(y=x^2+2\)
\(\,\,\,\,\,\,\,\) \(y=2x\)

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No solution

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\(\,\,\,\,\,x^2+2=2x\)

\(\,\,\,\,\,x^2-2x+2=0\)

\(\,\,\,\,\,\text{No real solutions}\)

\(\textbf{9)}\) \(y=x^2+2\)
\(\,\,\,\,\,\,\,\) \(y=x^2+4\)

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No solution

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\(\,\,\,\,\,x^2+2=x^2+4\)

\(\,\,\,\,\,2=4\)

\(\,\,\,\,\,\text{This is a contradiction, so there are no real solutions.}\)

\(\textbf{10)}\) \(\dfrac{1}{x-y}=\dfrac{2}{3x}\)

\(\,\,\,\,\,\,\,\,\,\,\) \(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{6}\)

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The solution is \( (18,-9) \)

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\(\,\,\,\,\,\dfrac{1}{x-y}=\dfrac{2}{3x}\)

\(\,\,\,\,\,3x=2(x-y)\)

\(\,\,\,\,\,3x=2x-2y\)

\(\,\,\,\,\,x=-2y\)

Now substitute \(x=-2y\) into the second equation:
\(\,\,\,\,\,\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{6}\)

\(\,\,\,\,\,\dfrac{1}{-2y}-\dfrac{1}{y}=\dfrac{1}{6}\)

\(\,\,\,\,\,-\dfrac{1}{2y}-\dfrac{1}{y}=\dfrac{1}{6}\)

\(\,\,\,\,\,-\dfrac{1+2}{2y}=\dfrac{1}{6}\)

\(\,\,\,\,\,-\dfrac{3}{2y}=\dfrac{1}{6}\)

\(\,\,\,\,\,2y\cdot\dfrac{1}{6}=-3\)

\(\,\,\,\,\,\dfrac{y}{3}=-3\)

\(\,\,\,\,\,y=-9\)

\(\,\,\,\,\,x=-2y=18\)

\(\,\,\,\,\,(18,-9)\)

\(\textbf{11)}\) *Challenge Problem*
\(\,\,\,\,\,\,\,\,\,\,\) \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\)

\(\,\,\,\,\,\,\,\,\,\,\) \(y=x-1\)

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The solutions are \(\left(\dfrac{9+12\sqrt{3}}{13},\,\dfrac{-4+12\sqrt{3}}{13}\right), \enspace \left(\dfrac{9-12\sqrt{3}}{13},\,\dfrac{-4-12\sqrt{3}}{13}\right)\)

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\(\,\,\,\,\,\dfrac{x^2}{9}+\dfrac{(x-1)^2}{4}=1\)

\(\,\,\,\,\,\Rightarrow \dfrac{4x^2}{36}+\dfrac{9(x-1)^2}{36}=\dfrac{36}{36}\)

\(\,\,\,\,\,\Rightarrow 13x^2-18x-27=0\)

\(\,\,\,\,\,x=\dfrac{18\pm\sqrt{(-18)^2-4(13)(-27)}}{2\cdot 13}\)

\(\,\,\,\,\,=\dfrac{18\pm 24\sqrt{3}}{26}=\dfrac{9\pm 12\sqrt{3}}{13}\)

\(\,\,\,\,\,y=x-1=\dfrac{-4\pm 12\sqrt{3}}{13}\)

\(\,\,\,\,\,\left(\dfrac{9+12\sqrt{3}}{13},\dfrac{-4+12\sqrt{3}}{13}\right),\ \left(\dfrac{9-12\sqrt{3}}{13},\dfrac{-4-12\sqrt{3}}{13}\right)\)

\(\textbf{12)}\) \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\)

\(\,\,\,\,\,\,\,\,\,\,\) \(y=2x\)

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No solution

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\(\,\,\,\,\,\dfrac{x^2}{16}-\dfrac{(2x)^2}{9}=1\)

\(\,\,\,\,\,\Rightarrow \dfrac{x^2}{16}-\dfrac{4x^2}{9}=1\)

\(\,\,\,\,\,\Rightarrow \dfrac{9x^2-64x^2}{144}=1\)

\(\,\,\,\,\,\Rightarrow -\dfrac{55x^2}{144}=1\)

\(\,\,\,\,\,\Rightarrow x^2=-\dfrac{144}{55}\ \text{(no real }x\text{)}\)

\(\,\,\,\,\,\text{Therefore, no real intersection.}\)

\(\textbf{13)}\) \(x^2+y^2=20\)
\(\,\,\,\,\,\,\,\,\,\,\) \(y=|x|\)

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The solutions are \( (\sqrt{10},\sqrt{10}), \enspace (-\sqrt{10},\sqrt{10}) \)

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\(\,\,\,\,\,x^2+y^2=20 \text{ and } y=|x|\)

\(\,\,\,\,\,x^2+(|x|)^2=20\)

\(\,\,\,\,\,2x^2=20\)

\(\,\,\,\,\,x^2=10\)

\(\,\,\,\,\,x=\pm\sqrt{10}\)

\(\,\,\,\,\,y=|x|=\sqrt{10}\)

\(\,\,\,\,\,(\sqrt{10},\sqrt{10}),(-\sqrt{10},\sqrt{10})\)

\(\textbf{14)}\) \((x-2)^2+(y+3)^2=16\)
\(\,\,\,\,\,\,\,\,\,\,\) \(y=-1\)

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The solutions are \( (2+2\sqrt{3},-1), \enspace (2-2\sqrt{3},-1) \)

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\(\,\,\,\,\,(x-2)^2+(y+3)^2=16 \text{ and } y=-1\)

\(\,\,\,\,\,(x-2)^2+(2)^2=16\)

\(\,\,\,\,\,(x-2)^2=12\)

\(\,\,\,\,\,x-2=\pm\sqrt{12}=\pm 2\sqrt{3}\)

\(\,\,\,\,\,x=2\pm 2\sqrt{3},\quad y=-1\)

\(\,\,\,\,\,(2+2\sqrt{3},-1),(2-2\sqrt{3},-1)\)

\(\textbf{15)}\) \(y=x^2\)
\(\,\,\,\,\,\,\,\,\,\,\) \(y=\dfrac{6}{x}\)

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The solution is \(\left(\sqrt[3]{6},\,\sqrt[3]{36}\right)\)

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\(\,\,\,\,\,x^2=\dfrac{6}{x}\quad (x\neq 0)\)

\(\,\,\,\,\,x^3=6\)

\(\,\,\,\,\,x=\sqrt[3]{6}\)

\(\,\,\,\,\,y=x^2=\left(\sqrt[3]{6}\right)^2=\sqrt[3]{36}\)

\(\,\,\,\,\,\left(\sqrt[3]{6},\sqrt[3]{36}\right)\)

Graph each system of inequalities.

\(\textbf{16)}\) \( y\gt x^2 \)
\(\,\,\,\,\,\,\,\,\,\,\) \( y\lt4 \)

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\(\textbf{17)}\) \( x^2+y^2\le36 \)
\(\,\,\,\,\,\,\,\,\,\,\) \( y\ge3 \)

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\(\textbf{18)}\) \( y\ge x^2-4 \)
\(\,\,\,\,\,\,\,\,\,\,\) \( y\le -x^2+6 \)

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\(\textbf{19)}\) \( y\le x^2+3 \)
\(\,\,\,\,\,\,\,\,\,\,\) \( y\ge x^2-5 \)

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Related Pages\(\)

\(\bullet\text{ Solving Systems with Substitution}\)
\(\,\,\,\,\,\,\,\,y=−2x+5\)
\(\,\,\,\,\,\,\,\,y=x-1…\)

\(\bullet\text{ Solving Systems with Elimination}\)
\(\,\,\,\,\,\,\,\,7x+4y=31 \)
\(\,\,\,\,\,\,\,\,3x+2y=15…\)

\(\bullet\text{ Graphing Systems of Inequalities}\)
\(\,\,\,\,\,\,\,\,y\lt-3x+2 \)
\(\,\,\,\,\,\,\,\,y\ge\frac{1}{2}x-1…\)

\(\bullet\text{ 3 variable systems}\)
\(\,\,\,\,\,\,\,\,2x+3y−5z=−7 \)
\(\,\,\,\,\,\,\,\,3x−6y+4z=3 \)
\(\,\,\,\,\,\,\,\,x+4y+2z=15…\)

\(\bullet\text{ Andymath Homepage}\)

Frequently Asked Questions

What does it mean to solve a non-linear system of equations?
Solving a non-linear system means finding the set of variable values that satisfy all equations in the system at the same time. Unlike linear systems, these involve at least one non-linear expression, such as a quadratic, circle, parabola, hyperbola, exponential, or radical.

What methods can be used to solve non-linear systems?
\(\,\,\,\,\,\cdot\) Substitution – replace one variable from one equation into the other.
\(\,\,\,\,\,\cdot\) Elimination – combine equations to eliminate a variable.
\(\,\,\,\,\,\cdot\) Graphing – plot the equations and identify intersection points.
\(\,\,\,\,\,\cdot\) Technology – graphing calculators, Desmos, or algebra software for complex systems.

Why is it important to check solutions?
Because operations like squaring, taking square roots, and dealing with absolute values can introduce extraneous solutions. Always substitute your results back into the original equations to confirm they work.

How many solutions can a non-linear system have?
It can have 0 (no intersection), 1 (tangent point), 2 or more (multiple intersections), or infinitely many if the equations describe the same curve.

Why are non-linear systems important in math and real life?
They appear in physics (projectile motion vs. linear paths), economics (supply and demand curves), engineering (stress-strain models), and biology (population growth). They also build the skill of analyzing where different models intersect, which is foundational in algebra and calculus.

What are common mistakes to avoid?
\(\,\,\,\,\,\cdot\) Forgetting to check for extraneous solutions.
\(\,\,\,\,\,\cdot\) Confusing approximate graphing answers with exact algebraic answers.
\(\,\,\,\,\,\cdot\) Stopping after one solution when multiple intersections exist.

At what level of math are non-linear systems usually taught?
They are typically introduced in Algebra II or College Algebra, and appear again in Precalculus and Calculus.

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