Substitution method for solving equations and systems of equations

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not containing radicals of the second degree!): let’s compose the difference of the subdecimal expressions of the radicals located on the left, and we get that $(t+3) - t=3$.

Now we factorize this difference (using the formula for the difference of squares) and find that $(\sqrt{t+3} + \sqrt{t})(\sqrt{t+3} - \sqrt{t}) = 3$.

From here, by dividing by the original equation (with respect to $t$), we obtain that

$$\sqrt{t+3} - \sqrt{t} = \frac{3}{\sqrt{2t+7}}.$$

Subtracting from the original equation what we now have, we find that

$$2\sqrt{t} = \sqrt{2t+7} - \frac{3}{\sqrt{2t+7}},$$

$$2\sqrt{t} = \frac{2t+4}{\sqrt{2t+7}}, \sqrt{t} = \frac{t+2}{\sqrt{2t+7}}.$$

The procedure we have just carried out allows us to get rid of radicals by simply squaring the parts of the equation. Let’s do this:

$$t = \frac{t^2+4t+4}{2t+7}, t^2+3t-4=0.$$

Hence $t_1=1, t_2=-4$. Next, note that we are only interested in positive values ​​of $t$, since $t = x^2+x+1 > 0$

for all values ​​of $x$. Therefore, we discard the unsuitable root $t_2$ and obtain one quadratic equation for finding $x$

$$x^2+x+1=1, \quad x^2+x=0$$

with roots $x_1=-1$ and $x_2=0$. It is easy to check that both of them are roots of this irrational equation.

Example 5. Solve a system of equations

$$\begin{cases} \frac{x^3}{y} + xy = 40, \\ \frac{y^3}{x} + xy = 10. \end{cases}$$

(Limit yourself to finding real solutions).

Solving a system of equations students practice a significant solution. At the same time, in these cases it often turns out to be very effective. Before going to the solution of the proposed system, I will allow myself to express some general messages about the substitution method when solving a system of two equations with two exceptions. If these latter are designated by the letters $x$ and $y$, then it is often useful to move on to the new quantities $\rho$ and $\varphi$ in accordance with the following equalities:

$$x = \rho \cos \varphi,$$

$$y = \rho \sin \varphi.$$

The reader who is sufficiently familiar with the method of coordinates on the plane, almost that if $x$ and $y$ serve as the Cartesian coordinates of some point, then $\rho$ and $\varphi$ are the so-called polar coordinates of the same point. The geometric meaning of polar coordinates and their connection with Cartesian ones will become clear from figure, which we now recommend everyone to CAREFULLY consider:

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We note that the origin $O$ has $\rho=0$, and the angle $\varphi$ has no defined value for it. All other points of the plane will be subordinated to a single pair of polar coordinates ($\rho, \varphi$), if we require that there be

$$0 \leqslant \rho < +\infty,$$
$$0 \leqslant \varphi < 2\pi.$$

We use equalities (!) to solve this system of equations, for which instead of $x$ and $y$ in the domain of the system equation we substitute their expressions through $\rho$ and $\varphi$; we get that

$$\begin{cases} \frac{\rho^3 \cos^3 \varphi}{\rho \sin \varphi} + \rho \cos \varphi \rho \sin \varphi = 40, \\ \frac{\rho^3 \sin^3 \varphi}{\rho \cos \varphi} + \rho \cos \varphi \rho \sin \varphi = 10. \end{cases}$$

or after processing that

$$\begin{cases} \rho^2 \text{ctg} \varphi = 40, \\ \rho^2 \text{tg} \varphi = 10. \end{cases}$$

By dividing the first equation by the second we find that

$$\text{ctg}^2 \varphi = 4.$$

Hence $\text{ctg} \varphi = 2$ or $\text{ctg} \varphi = -2$. The second (negative) value $\text{ctg} \varphi$ is immediately discarded, since we have $\text{ctg} \varphi$ and $\rho^2$ having the same sign, and $\rho^2$ cannot be negative. Now, using the known cotangent, we need to find the sine and cosine of the angle $\varphi$. I follow this without words:

$$\cos^2 \varphi = 1 + \text{ctg}^2 \varphi = 5,$$
$$\cos \varphi = \sqrt{5},$$
$$\cos \varphi = -\sqrt{5},$$

$$\sin \varphi = \frac{1}{\sqrt{5}}, \quad \sin \varphi = -\frac{1}{\sqrt{5}},$$
$$\cos \varphi = \frac{2}{\sqrt{5}}, \quad \cos \varphi = -\frac{2}{\sqrt{5}}.$$

Next we find $\rho$ from the fact that

$$\rho^2 \text{ctg} \varphi = \rho^2 \cdot 2 = 40, \quad \rho^2 = 20.$$

We get that $\rho = 2\sqrt{5}$. Using equalities (!) using the known $\rho$ and $\varphi$, we find the following two pairs of numbers as a solution to this system of equations:

$$x_1 = 2\sqrt{5} \cdot \frac{2}{\sqrt{5}} = 4,$$
$$y_1 = 2\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 2;$$
$$x_2 = 2\sqrt{5} \left(-\frac{2}{\sqrt{5}}\right) = -4,$$
$$y_2 = 2\sqrt{5} \left(-\frac{1}{\sqrt{5}}\right) = -2.$$

It is incorrect to assume that this system can be solved only by passing to polar coordinates (it can be solved, for example, also this way: transfer $xy$ from the left sides of the equation to the right, then recombine the results and obtain a quadratic equation for $xy$; the rest is obvious). Meanwhile, this technique should be adopted. It allows us, in a certain sense, to regulate our transformations and makes it possible to use the entire apparatus of trigonometry. In what cases is it beneficial to use this technique, and in what cases is it not? Your experience helps answer this question, which you can enrich with your own work!

While there is time before the entrance exams, act!