8. Non-linear systems · Equations and systems

What is a non-linear system

Linear vs. non-linear

A linear system of 2 equations in two unknowns is made up of equations of degree 1. For example:

\begin{cases} 2x - 3y = 5, \\ x + 2y = 3. \end{cases}

It is solved by elimination, equating or substitution — any of the three methods always works.

A non-linear system is one where at least one equation is not of degree 1: products $xy$, squares, roots, exponentials, logarithms, etc. may appear.

Strategy: substitution with eyes open

The main method is substitution: we isolate one unknown in the "simpler" equation and plug it into the other. But, unlike linear systems, non-linear systems have no single, fixed method — each system has its own structure and it pays to look at it before starting.

Useful tools that will show up in the examples:

· Change of variable: $\,t = \log x,\; z = \log y\,$ turns a logarithmic system into a linear system in $t, z$.

· Exponential substitution: $\,t = 2^x,\; z = 2^y\,$ turns a system of exponentials into a system in $t, z$.

· Special products and elimination: by combining the equations with suitable signs, $(x-y)^{2}$ or $(x+y)^{2}$ may appear.

· Vieta: if we have a sum and a product ($\,a+b = S,\; a\cdot b = P\,$), then $a$ and $b$ are the roots of $w^{2} - Sw + P = 0$.

Final check

In systems with logarithms or radicals, some candidate values may fall outside the domain ($\log$ of a number $\leq 0$, $\sqrt{\cdot}$ of a negative). Always substitute back to discard extraneous solutions.

Introductory example — two ways

$\begin{cases} xy = -6 \\ x^{2} + y^{2} = 13 \end{cases}$

It has a product $xy$ and a sum of squares: clearly not linear. We can tackle it two ways.

WAY 1 Direct substitution. From the first, $\,y = -\dfrac{6}{x}\,$ (assuming $x \neq 0$). In the second:

x^{2} + \dfrac{36}{x^{2}} = 13.

Multiply by $x^{2}$: $\;x^{4} - 13x^{2} + 36 = 0$ (biquadratic). Substitute $t = x^{2}$:

t^{2} - 13t + 36 = 0 \;\Longrightarrow\; t = \dfrac{13 \pm 5}{2} \;\Rightarrow\; t = 9\ \text{or}\ t = 4.

So $x = \pm 3$ or $x = \pm 2$, and $y = -6/x$ gives respectively $y = \mp 2$ or $y = \mp 3$.

WAY 2 Combining into a special product. Multiply the first by $-2$ and add the second:

\begin{aligned} -2xy &= 12, \\ x^{2} + y^{2} &= 13. \end{aligned} \quad\Longrightarrow\quad x^{2} - 2xy + y^{2} = 25 \;\Longleftrightarrow\; (x-y)^{2} = 25.

Hence $\,x - y = \pm 5$.

· If $x - y = 5$: $\,x = y + 5\,$ and $\,xy = -6 \Rightarrow (y+5)y = -6 \Rightarrow y^{2} + 5y + 6 = 0 \Rightarrow y = -2$ or $y = -3$. Solutions: $(3, -2)$ and $(2, -3)$.

· If $x - y = -5$: $\,x = y - 5\,$ and $\,(y-5)y = -6 \Rightarrow y^{2} - 5y + 6 = 0 \Rightarrow y = 2$ or $y = 3$. Solutions: $(-3, 2)$ and $(-2, 3)$.

Both ways agree on four solutions:

\boxed{\;(x,y) \in \{(3,-2),\ (-3,2),\ (2,-3),\ (-2,3)\}.\;}

Way 2 is more elegant here because it exploits the symmetry of $x^{2}+y^{2}$ with $-2xy$. Way 1 always works, but it leads to a biquadratic and four branches.

Systems with two logarithmic equations

Recipe

When both equations are sums and differences of logarithms (or become so after applying the properties $\log(BC) = \log B + \log C$ and $\log(B/C) = \log B - \log C$), we make the substitution:

t = \log x, \qquad z = \log y.

The system becomes linear in $t, z$, we solve it as usual, and at the end we undo the substitution: $x = 10^{t}$, $y = 10^{z}$.

Example b) $\begin{cases} 2\log x + \log y = 5 \\ \log(xy) = 4 \end{cases}$

We rewrite the second: $\log x + \log y = 4$. With the substitution $t = \log x$, $z = \log y$:

\begin{cases} 2t + z = 5, \\ t + z = 4. \end{cases}

By elimination: $t = 1$, and from the second $z = 3$. Undoing the substitution: $x = 10^{1} = 10$, $y = 10^{3} = 1000$.

Check: $2\log 10 + \log 1000 = 2 + 3 = 5$ ✓; $\log(10 \cdot 1000) = \log 10000 = 4$ ✓.

\boxed{\;x = 10,\ y = 1000.\;}

Example f) $\begin{cases} \log(xy) = 4 \\ \log\!\dfrac{x}{y^{2}} = -2 \end{cases}$

We apply the properties: $\log x + \log y = 4$ and $\log x - 2\log y = -2$. With the substitution $t = \log x$, $z = \log y$:

\begin{cases} t + z = 4, \\ t - 2z = -2. \end{cases}

Subtracting: $3z = 6 \Rightarrow z = 2$, and $t = 4 - 2 = 2$. So $x = 100$, $y = 100$.

Check: $\log(100 \cdot 100) = 4$ ✓; $\log\bigl(\tfrac{100}{10000}\bigr) = \log 0.01 = -2$ ✓.

\boxed{\;x = 100,\ y = 100.\;}

Example m) $\begin{cases} \log x + 3\log y = 5 \\ \log\!\dfrac{x^{2}}{y} = 3 \end{cases}$

The second gives $2\log x - \log y = 3$. Substitute $t = \log x$, $z = \log y$:

\begin{cases} t + 3z = 5, \\ 2t - z = 3 \;\Rightarrow\; z = 2t - 3. \end{cases}

Substituting: $t + 3(2t - 3) = 5 \Rightarrow 7t = 14 \Rightarrow t = 2$, and $z = 1$. So $x = 100$, $y = 10$.

Check: $\log 100 + 3\log 10 = 2 + 3 = 5$ ✓; $\log(10000/10) = \log 1000 = 3$ ✓.

\boxed{\;x = 100,\ y = 10.\;}

Mixed: one linear + one logarithmic

Example g) $\begin{cases} x - 5y = -5 \\ \log x + \log y = 1 \end{cases}$

Here we do not need the substitution $t = \log x$ — it is shorter to isolate from the linear equation.

WAY 1 From the linear equation: $x = 5y - 5$. We substitute into the second, after combining the logarithms:

\log\!\bigl[(5y-5)\,y\bigr] = 1 \;\Longrightarrow\; (5y-5)y = 10 \;\Longrightarrow\; 5y^{2} - 5y - 10 = 0.

y^{2} - y - 2 = 0 \;\Longrightarrow\; (y-2)(y+1) = 0 \;\Rightarrow\; y = 2\ \text{or}\ y = -1.

We discard $y = -1$ (we cannot take $\log(-1)$). For $y = 2$: $\,x = 5\cdot 2 - 5 = 5$.

WAY 2 If we prefer, we can work the logarithmic equation first: $\log(xy) = 1 \Rightarrow xy = 10$. The system becomes $\{x - 5y = -5,\; xy = 10\}$, which is also solved by substitution and leads to the same quadratic.

Check: $5 - 5\cdot 2 = -5$ ✓; $\log 5 + \log 2 = \log 10 = 1$ ✓.

\boxed{\;x = 5,\ y = 2.\;}

Radical + exponential / logarithmic

Example d) $\begin{cases} \sqrt{x-3} = y - 2 \\ 2^{x} = 4^{y-1} \end{cases}$

We start with the exponential equation, which linearizes easily: $\,2^{x} = 4^{y-1} = 2^{2(y-1)} \Rightarrow x = 2y - 2$.

We substitute into the root: $\sqrt{2y-2-3} = y - 2 \Rightarrow \sqrt{2y-5} = y - 2$. Square both sides:

2y - 5 = (y-2)^{2} = y^{2} - 4y + 4 \;\Longrightarrow\; 0 = y^{2} - 6y + 9 = (y-3)^{2}.

Double root $y = 3$, and $x = 2\cdot 3 - 2 = 4$.

Check: $\sqrt{4-3} = 1 = 3 - 2$ ✓; $2^{4} = 16 = 4^{2} = 4^{3-1}$ ✓.

\boxed{\;x = 4,\ y = 3.\;}

Example l) $\begin{cases} \sqrt{x-1} = y + 2 \\ \log x - \log y = 1 \end{cases}$

WAY 1 We start with the logarithmic equation: $\log\!\dfrac{x}{y} = 1 \Rightarrow \dfrac{x}{y} = 10 \Rightarrow x = 10y$. Substituting:

\sqrt{10y - 1} = y + 2 \;\Longrightarrow\; 10y - 1 = y^{2} + 4y + 4.

0 = y^{2} - 6y + 5 = (y-1)(y-5) \;\Rightarrow\; y = 1\ \text{or}\ y = 5.

So $(x, y) = (10, 1)$ or $(50, 5)$.

WAY 2 Alternative: isolate $x$ from the root. $\sqrt{x-1} = y+2 \Rightarrow x = (y+2)^{2} + 1 = y^{2} + 4y + 5$. Substituting into the logarithmic equation: $\log(y^{2}+4y+5) - \log y = 1 \Rightarrow \log\!\dfrac{y^{2}+4y+5}{y} = 1$, from which $y^{2} + 4y + 5 = 10y$, i.e. the same quadratic as in way 1.

Check: $(10, 1)$: $\sqrt{9} = 3 = 1+2$ ✓, $\log 10 - \log 1 = 1$ ✓. $(50, 5)$: $\sqrt{49} = 7 = 5+2$ ✓, $\log 50 - \log 5 = \log 10 = 1$ ✓.

\boxed{\;(x, y) \in \{(10, 1),\ (50, 5)\}.\;}

Systems with two exponentials

Example h) $\begin{cases} 3 \cdot 2^{x} - 2 \cdot 3^{y} = 6 \\ \log_{2}(3^{y} - 1) = x \end{cases}$

From the second, by the inverse definition of the logarithm: $\,3^{y} - 1 = 2^{x}\,$, that is, $\,2^{x} = 3^{y} - 1$. We substitute into the first:

3(3^{y} - 1) - 2 \cdot 3^{y} = 6 \;\Longrightarrow\; 3 \cdot 3^{y} - 3 - 2 \cdot 3^{y} = 6 \;\Longrightarrow\; 3^{y} = 9 = 3^{2}.

So $y = 2$, and $\,2^{x} = 3^{2} - 1 = 8 = 2^{3} \Rightarrow x = 3$.

Check: $3 \cdot 8 - 2 \cdot 9 = 24 - 18 = 6$ ✓; $\log_{2}(9-1) = \log_{2} 8 = 3$ ✓.

\boxed{\;x = 3,\ y = 2.\;}

Example n) $\begin{cases} 2^{x} + 2^{y} = 12 \\ 2^{x} \cdot 2^{y} = 32 \end{cases}$

Here we have a sum and a product of two quantities — this invites the use of Vieta.

WAY 1 Change of variable + Vieta. Let $\,t = 2^{x},\; z = 2^{y}$. Then:

t + z = 12, \qquad t \cdot z = 32.

$t$ and $z$ are the roots of $\,w^{2} - 12w + 32 = 0$:

w = \dfrac{12 \pm \sqrt{144-128}}{2} = \dfrac{12 \pm 4}{2} \;\Rightarrow\; w = 8\ \text{or}\ w = 4.

So $\{t, z\} = \{8, 4\}$, that is, $\,2^{x} = 8 = 2^{3},\; 2^{y} = 4 = 2^{2}\,$ (or the other way round). Solutions: $(3, 2)$ and $(2, 3)$.

WAY 2 Product of exponentials → sum of exponents. $\,2^{x} \cdot 2^{y} = 2^{x+y} = 32 = 2^{5} \Rightarrow x + y = 5$. Then $y = 5 - x$, and the first gives:

2^{x} + 2^{5-x} = 12 \;\Longleftrightarrow\; 2^{x} + \dfrac{32}{2^{x}} = 12.

Multiplying by $2^{x}$ and with the substitution $t = 2^{x}$: $\,t^{2} - 12t + 32 = 0$ — the same quadratic as in way 1. We recover the same two solutions.

Check: $(3, 2)$: $\,8 + 4 = 12$ ✓ and $\,8 \cdot 4 = 32$ ✓. $(2, 3)$: symmetric ✓.

\boxed{\;(x, y) \in \{(3, 2),\ (2, 3)\}.\;}