Nonlinear Systems of Equations
Definition of Nonlinear Systems of Equations
A system of equations in which any one equation is not linear is a nonlinear system. A nonlinear system may have no solutions, an infinite set of solutions, or any number of real or complex solutions.
Solutions of Nonlinear Systems in Two Variables
Solutions of nonlinear systems in two variables can be found by three methods:
GRAPHICAL METHOD. Graph each equation. The coordinates of any points of intersection may be read from the graph. After checking by substitution in each equation, these coordinates are the real solutions of the system. Normally, only approximations to real solutions can be found by this method, but when the algebraic methods below fail, this method can still be used.
SUBSTITUTION METHOD. Solve one equation for one variable in terms of the other. Substitute this expression into the other equations to determine the value of the first variable (if possible). Then substitute this value to determine the value of the other variable.
ELIMINATION METHOD. Apply the operations on equations leading to equivalent systems to eliminate one variable from one equation, solve the resulting equation for this variable, and substitute this value to determine the value of the other variable.
$ graphically.
The graph of $ y = e^{-x} $ is an exponential decay curve; the graph of $ y = 1 + x $ is a straight line.
Sketch the two graphs in the same coordinate system (see Fig. 33-1).
The graphs appear to intersect at $ (0,1) $ . Substituting x=0, y=1 into $ y=e^{-x} $ yields $ 1=e^{-0} $ or 1=1. Substituting into $ y=1+x $ yields $ 1=1+0 $ . Thus, $ (0,1) $ is a solution of the system. The method does not rule out the possibility of other solutions, including nonreal complex solutions.
EXAMPLE 33.2 Solve the system $ x + 2y = 11 $ (1) by substitution.
Substitute the expression $ x^{2} - 2 $ from equation (1) into equation (2) for y to obtain
\[ x+2(x^{2}-2)=11 \]
Solving this quadratic equation in x yields
\[ 2x^{2}+x-15=0 \]
\[ (2x-5)(x+3)=0 \]
\[ \begin{aligned}2x-5&=0\quad&or\quad&x+3&=0\\x&=\frac{5}{2}\quad&x&=-3\end{aligned} \]
Substituting these values for x into equation (1) yields
\[ x=\frac{5}{2}:y=\left(\frac{5}{2}\right)^{2}-2=\frac{17}{4}\quad x=-3:y=(-3)^{2}-2=7 \]
Thus the solutions are $ (,) $ and $ (-3,7) $ .
EXAMPLE 33.3 Solve by elimination: $ \[\begin{aligned}x^{2} + y^{2} &= 1 \quad (1)\\x^{2} - y^{2} &= 7 \quad (2)\end{aligned}\]$
Replacing equation (2) by itself plus equation (1) yields the equivalent system:
\[ \begin{aligned}x^{2}+y^{2}&=1\quad&(1)\\2x^{2}&=8\quad&(3)\end{aligned} \]
Solving equation (3) for x yields
\[ x^{2}=4 \]
\[ x=2\qquad or\qquad x=-2 \]
Substituting these values for x into equation (1) yields
\[ \begin{align*}x=2:\quad&2^{2}+y^{2}=1\quad&x=-2:\quad&(-2)^{2}+y^{2}=1\\y^{2}=&-3\quad&y^{2}=&-3\\y=i\sqrt{3}\quad&or\quad y=-i\sqrt{3}\quad&y=i\sqrt{3}\quad&or\quad y=-i\sqrt{3}\end{align*} \]
Thus the solutions are \((2, i\sqrt{3})\), \((2, -i\sqrt{3})\), \((-2, i\sqrt{3})\), \((-2, -i\sqrt{3})\).
No General Procedure Exists
There is no general procedure for solving nonlinear systems of equations. Sometimes a combination of the above methods is effective; frequently no algebraic method works and the graphical method can be used to find some approximate solutions, which can then be refined by advanced numerical methods.
SOLVED PROBLEMS
33.1. Solve the system $ \[\begin{array}{r} y = x^{2} \\ x + y = 2 \end{array}\]$ and illustrate graphically. (1)
Solve by substitution: Substitute the expression $ x^{2} $ from equation (1) into equation (2) for y to obtain the quadratic equation
\[ x+x^{2}=2 \]
Solving yields
\[ \begin{aligned}x^{2}+x-2&=0\ $ x-1)(x+2)&=0\\x&=1\quad\quad or\quad\quad x=-2\end{aligned} \]
Substituting these values for x into equation (1) yields:
\[ x=1:y=1^{2}=1\qquad x=-2:y=(-2)^{2}=4 \]
Thus the solutions are $ (1, 1) $ and $ (-2, 4) $ .
The graph of $ y = x^{2} $ is the basic parabola, opening up. The graph of $ x + y = 2 $ is a straight line with slope -1 and y intercept 2. Sketch the two graphs in the same coordinate system (Fig. 33-2).
$ (1) and illustrate graphically. (2)
Solve by substitution: Substitute the expression $ x^{2} + 2 $ from equation (1) into equation (2) for y to obtain the quadratic equation
\[ x^{2}+2=2x-4 \]
Solving yields
\[ \begin{aligned}x^{2}-2x+6&=0\\x&=\frac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(6)}}{2(1)}\\x&=1\pm i\sqrt{5}\end{aligned} \]
Substituting these values for x into equation (1) yields:
\[ \begin{aligned}&x=1+i\sqrt{5}:y=(1+i\sqrt{5})^{2}+2=-2+2i\sqrt{5}\\&x=1-i\sqrt{5}:y=(1-i\sqrt{5})^{2}+2=-2-2i\sqrt{5}\\ \end{aligned} \]
Thus the solutions are $ (1 + i, -2 + 2i) $ and $ (1 - i, -2 - 2i) $ .
The graph of $ y = x^{2} + 2 $ is the basic parabola, opening up, shifted up 2 units. The graph of y = 2x - 4 is a straight line with slope 2 and y intercept -4. Sketch the two graphs in the same coordinate system and note that the complex solutions correspond to the fact that the graphs do not intersect (Fig. 33-3).
$ (1)
The system is most efficiently solved by elimination. Replace equation (2) by itself plus four times equation (1):
\[ \begin{aligned}&4y^{2}-16x^{2}=16\quad&4\cdot Eq.(1)\\&\frac{9y^{2}+16x^{2}=140}{13y^{2}}&=156\quad&(2)\\\end{aligned} \]
Solving equation (3) yields
\[ \begin{aligned}&y^{2}=12\\&y=\pm2\sqrt{3}\\ \end{aligned} \]
Substituting these values for y into equation (1) yields
\[ \begin{align*}y=2\sqrt{3}:\quad&(2\sqrt{3})^{2}-4x^{2}=4\quad&y=-2\sqrt{3}:\quad&(-2\sqrt{3})^{2}-4x^{2}=4\\&x^{2}=2\quad&x^{2}=2\\x=\sqrt{2}\quad&or\quad x=-\sqrt{2}\quad&x=\sqrt{2}\quad&or\quad x=-\sqrt{2}\end{align*} \]
Thus the solutions are \((\sqrt{2}, 2\sqrt{3})\), \((\sqrt{2}, -2\sqrt{3})\), \((-\sqrt{2}, 2\sqrt{3})\), \((-\sqrt{2}, -2\sqrt{3})\).
33.4. Solve the system $ \[\begin{array}{r}x^{2}+xy-3y^{2}=3\\x^{2}+4xy+3y^{2}=0\end{array}\]$ (1)
The system is most efficiently solved by substitution. Solve equation (2) for x in terms of y:
\[ \begin{aligned}(x+y)(x+3y)&=0\\x+y&=0\quad\quad or\quad\quad x+3y=0\\x&=-y\quad\quad\quad\quad x=-3y\end{aligned} \]
Now substitute these expressions for x into equation (1):
\[ \begin{aligned}If x=-y:(-y)^{2}+(-y)y-3y^{2}&=3\\-3y^{2}&=3\\y&=i\quad\quad or\quad\quad y=-i\end{aligned} \]
Since \(x = -y\), when \(y = i\), \(x = -i\), and when \(y = -i\), \(x = i\).
\[ \begin{aligned}If x=-3y:(-3y)^{2}+(-3y)y-3y^{2}&=3\\3y^{2}&=3\\y&=1\quad\quad or\quad\quad y=-1\end{aligned} \]
Since x = -3y, when y = 1, x = -3, and when y = -1, x = 3.
Thus the solutions are $ (i,-i) $ , $ (-i,i) $ , $ (-3,1) $ , $ (3,-1) $ .
33.5. Solve the system $ \[\begin{array}{r}x^{2}+xy-y^{2}=-1\\x^{2}+2xy-y^{2}=1\end{array}\]$ (1)
This can be solved by a combination of elimination and substitution techniques. Replace equation (2) by itself plus -1 times equation (1):
\[ \begin{aligned}\frac{-x^{2}-xy+y^{2}&=1\quad&(-1)\cdot Eq.(1)\\\frac{x^{2}+2xy-y^{2}&=1\quad&(2)\\xy&=2\quad&(3)\end{aligned} \]
Solving equation (3) for y in terms of x yields $ y = $ . Substitute the expression $ $ into equation (1) for y to obtain:
\[ \begin{aligned}x^{2}+x\Big(\frac{2}{x}\Big)-\Big(\frac{2}{x}\Big)^{2}&=-1\\x^{2}+2-\frac{4}{x^{2}}&=-1\\x^{2}+3-\frac{4}{x^{2}}&=0\\x^{4}+3x^{2}-4&=0\qquad(x\neq0)\ $ x-1)(x+1)(x-2i)(x+2i)&=0\\x=1\quad\quad or\quad x=-1\quad\quad or\quad x=2i\quad\quad or\quad x=-2i\end{aligned} \]
Since $ y = $ , when x = 1, $ y = = 2 $ and when x = -1, $ y = = -2 $ . Also, when $ x = 2i $ , $ y = = -i $ .
and when $ x = -2i $ , $ y = = i $ . Thus, the solutions are $ (1,2) $ , $ (-1,-2) $ , $ (2i,-i) $ , and $ (-2i,i) $ .
33.6. An engineer wishes to design a rectangular television screen that is to have an area of 220 square inches and a diagonal of 21 inches. What dimensions should be used?
Let \(x = \text{width}\) and \(y = \text{length}\) of the screen. Sketch a figure (see Fig. 33-4).
Since the area of the rectangle is to be 220 square inches,
\[ xy=220 \]
Since the diagonal is to be 21 inches, from the Pythagorean theorem,
\[ x^{2}+y^{2}=21^{2} \]
The system (1), (2) can be solved by substitution. Solving equation (1) for y in terms of x yields
\[ y=\frac{220}{x} \]
Substitute the expression $ $ for y into equation (2) to obtain:
\[ x^{2}+\left(\frac{220}{x}\right)^{2}=441 \]
\[ x^{2}+\frac{48400}{x^{2}}=441 \]
\[ x^{4}+48400=441x^{2}\qquad(x\neq0) \]
\[ x^{4}-441x^{2}+48400=0 \]
The last equation is quadratic in $ x^{2} $ , but not factorable. Use the quadratic formula to obtain:
\[ x^{2}=\frac{-(-441)\pm\sqrt{(-441)^{2}-4(1)(48400)}}{2(1)} \]
\[ =\frac{441\pm\sqrt{881}}{2} \]
\[ x=\sqrt{\frac{441\pm\sqrt{881}}{2}} \]
In the last step, only the positive square root is meaningful. Thus the two possible solutions are
\[ x=\sqrt{\frac{441+\sqrt{881}}{2}}\approx15.34\qquad and\qquad x=\sqrt{\frac{441-\sqrt{881}}{2}}\approx14.34 \]
Since y = 220/x, if x = 15.34, y = 14.34, and conversely. Hence the only solution is for the dimensions of the screen to be $ 14.34 $ inches.
SUPPLEMENTARY PROBLEMS
33.7. Solve the systems: (a) $ 2y = x^{2} $ ; (b) $ x^{2} - y^{2} = 2 $
Ans. (a) $ (0,0) $ , $ (2,2) $ ; (b) $ (2,) $ , $ (2,-) $ , $ (-1,i) $ , $ (-1,-i) $
33.8. Solve the systems (a) $ x^{2} + y^{2} = 16 $ ; (b) $ x^{2} + y^{2} = 8 $ ; $ y^{2} = 4 - x $ ; y - x = 4
Ans. (a) $ (4,0) $ , $ (-3,) $ , $ (-3,-) $ ; (b) $ (-2,2) $
33.9. Solve the systems: (a) $ x^{2} + 4y^{2} = 24 $ ; (b) $ x^{2} - 8y^{2} = 1 $ ; $ x^{2} - 4y = 0 $ ; $ x^{2} + 4y^{2} = 25 $
Ans. (a) $ (, 2) $ , $ (-, 2) $ , $ (2i, -3) $ , $ (-2i, -3) $ ;
\[ (b)(\sqrt{17},\sqrt{2}),(\sqrt{17},-\sqrt{2}),(-\sqrt{17},\sqrt{2}),(-\sqrt{17},-\sqrt{2}) \]
33.10. Solve and illustrate the solutions graphically: (a) $ y = x^{2} - 1 $ ; (b) $ y = x^{2} - 2 $
Ans. (a) Solutions: $ (-1,0) $ , $ (3,8) $ ; Fig. 33-5:
- Solutions: $ (-2,2) $ , $ (1,-1) $ ; Fig. 33-6
33.11. Solve the systems: (a) $ 2x + 3y + xy = 16 $ ; (b) $ 2x^{2} - 5xy + 2y^{2} = 0 $ ; $ xy - 5 = 0 $ ; $ 3x^{2} + 2xy - y^{2} = 15 $
Ans. (a) $ (,2),(3,) $ ; (b) $ (2,1),(-2,-1),(,2),(-,-2) $
33.12. A rectangle of perimeter 100 meters is to be constructed to have area 100 square meters. What dimensions are required?
Ans. $ 25 + $ by $ 25 - $ , or approximately $ 47.91 $ meters