Exercises 2

This page allows you to practice some exercises on Solving Equations. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 3.4 - Some Nonlinear Equations - Section 3.5 - Using Implication Arrows - Section 3.6 - Two Linear Equations in Two Unknowns

Question 6

Solve the following equations. (a) $x^3 (1+x^2)(1 - 2x) = 0$

(b) $\frac{x^2 + 1}{x(x+1)} = 0$

(c) $\frac{(x+1)^{1/3} - \frac{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}} = 0$

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(a) $x = 0$ and $x = 1/2$.

(b) No solutions.

(c) $x = -3/2$.

Question 7

Examine what conclusions can be drawn about the variables if (a) $z^2(z-a) = z^3(a+b), \quad a \neq 0$

(b) $(1 + \ell)mx = (1 + \ell)my$

(c) $\frac{\lambda}{1+\mu} = -\frac{\lambda}{1 - \mu^2}$

(d) $ab - 2b - \phi b (2-a) = 0$

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(a) $z = 0$ or $z = a/(1-a-b)$ for $a + b \neq 1$. For $a + b = 1$, the only solution is $z=0$.

(b) $\ell = -1$ or $\mu = 0$ or $x=y$.

(c) $\lambda = 0$ and $\mu \neq \pm 1$, or $\mu=2$.

(d) $a = 2$ or $b=0$ or $\phi = -1$.

Question 8

Using implication arrows, solve the following equations. (a) $\frac{(x+1)^2}{x(x-1)} + \frac{(x-1)^2}{x(x+1)} - 2\frac{3x + 1}{x^2 -1} = 0$

(b) $x + 2 = \sqrt{4x + 13}$

(c) $x^2 - 2|x| - 3 = 0$

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(a) We can exclude $x \in \{-1, 0, 1\}$, as any of these values render a division by zero, which is not defined. Multiplying each term by the common denominator $x(x-1)(x+1)$, we derive the only solution using the following equivalences $\frac{(x+1)^2}{x(x-1)} + \frac{(x-1)^2}{x(x+1)} - 2\frac{3x + 1}{x^2 - 1} = 0 \text{ with } x \notin \{-1,0,1\}$.

$\iff (x+1)^3 - (x-1)^3 - 2x(3x + 1) = 0 \text{ with } x \notin \{-1,0,1\}$.

$\iff 2x(x^2 - 3x + 2) = 0 \text { with } x \notin \{-1, 0, 1\}$.

$\iff 2x(x-1)(x-2) = 0 \text{ with } x \notin \{ -1, 0, 1 \}$.

$\iff x = 2$.

(b) Squaring both sides and rearranging yields $x + 2 = \sqrt{4x + 13} \implies (x+2)^2 = 4x + 13 \implies x^2 = 9 \implies x = \pm 3$. But note that, by non-negativity of a square root, $x + 2 = \sqrt{4x + 13} \implies x+ 2 \geq 0$. Therefore, the only solution is $x=3$.

(c) The equivalent equation $|x|^2 - 2|x| - 3 = (|x| - 3)(|x| + 1) = 0$ gives $|x| = 3$ or $|x| = -1$. However, because $|x| \geq 0$, only $x = \pm 3$ are feasible solutions.

Question 9

Consider the following attempt to solve the equation $x + \sqrt{x + 4} = 2$:

"From the given equation, it follows that $\sqrt{x+4} = 2 - x$. Squaring both sides gives $x + 4 = 4 - 4x + x^2$. Rearranging terms shows that this equation implies $x^2 - 5x = 0$. Cancelling $x$, we obtain $x-5 = 0$, which is satisfied when $x=5$."

Mark with arrows the implications or equivalences expressed in the text. Which ones are correct? What is/are the correct solution(s) to the equation?

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$x + \sqrt{x + 4} = 2 \implies \sqrt{x+4} = 2 - x \implies x+4 = (2-x)^2$

$\implies x+4 = 4 - 4x + x^2 \implies x^2 - 5x = 0 \overset{(i)}{\implies} x-5 = 0 \overset{(ii)}{\impliedby} x = 5$

Note that

• Implication (i) is incorrect: $x^2 - 5x = 0 \implies x-5 =0$ or $x=0$. Note that the implication is formed by dividing by $x$, but this can only be done when $x \neq 0$.

• Implication (ii) is correct, but it breaks the chain of implications.

• Only $x=0$ is a correct solution. After correcting implication (i), we see that the given equation implies $x=0$ or $x=5$. But note that $x=5$ does not solve the original equation, it solves the different equation $x-\sqrt{x+4} = 2$ (i.e. with a minus instead of a plus).

Question 10

Solve the following systems of equations.

(a) $4x - 3y = 1 \text{ and } 2x + 9y = 4$

(b) $5x + 2y = 3 \text{ and } 2x + 3y = -1$

(c) $0.01r + 0.21s = 0.042 \text{ and } -0.25r + 0.55s = -0.47$

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(a) $x = 1/2, y = 1/3$.

(b) $x = 1, y= -1$.

(c) $r = 2.1, s= 0.1$.

Question 11

(a) Find two numbers whose sum is $52$ and whose difference is $26$.

(b) Five tables and $20$ chairs cost $1800$ Euros, whereas two tables and three chairs cost $420$ Euros. What is the price of each table and each chair?

(c) A firm produces headphones in two qualities, Basic (B) and Premium (P). For the coming year, the estimated output of B is $50\%$ higher than that of P. The profit per unit sold is $300$ Euros for P and $200$ Euros for B. If the profit target is $180.000$ Euros over the next year, how much of each of the two qualities must be produced?

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(a) $39$ and $13$.

(b) $120$ euros for a table and $60$ euros for a chair.

(c) $450$ of quality B and $300$ of quality P.