Ejercicio 184 – Solucionario Baldor

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DESCARGA

CAPITULO XXIV

Ecuaciones simultaneas con dos incognitas

Ejercicio 184

Resolver por determinantes:

$\begin{matrix} {\begin{matrix} 7 x + 8 y = 29 \\ 5 x + 11 y = 26 \end{matrix} \\ x = \frac{| \begin{matrix} 29 & 8 \\ 26 & 11 \end{matrix} |}{| \begin{matrix} 7 & 8 \\ 5 & 11 \end{matrix} |} & = \frac{29 (11) - 8 (26)}{7 (11) - 5 (8)} = \frac{319 - 208}{77 - 40} = \frac{}{37} = 3 \\ y = \frac{| \begin{matrix} 7 & 29 \\ 5 & 26 \end{matrix} |}{| \begin{matrix} 7 & 8 \\ 5 & 11 \end{matrix} |} & = \frac{7 (26) - 29 (5)}{7 (11) - 5 (8)} = \frac{182 - 145}{77 - 40} = \frac{37}{37} = 1 \\ S o l {\begin{matrix} x = 3 \\ y = 1 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 3 x - 4 y = 13 \\ 8 x - 5 y = - 5 \end{matrix} \\ x = \frac{| \begin{matrix} 13 & - 4 \\ - 5 & - 5 \end{matrix} |}{| \begin{matrix} 3 & - 4 \\ 8 & - 5 \end{matrix} |} & = \frac{13 (- 5) - (- 5) (- 4)}{3 (- 5) - 8 (- 4)} = \frac{- 65 - 20}{- 15 + 32} = \frac{-}{17} = - 5 \\ y = \frac{| \begin{matrix} 3 & 13 \\ 8 & - 5 \end{matrix} |}{| \begin{matrix} 3 & - 4 \\ 8 & - 5 \end{matrix} |} & = \frac{(- 5) (3) - 13 (8)}{3 (- 5) - 8 (- 4)} = \frac{- 15 - 104}{- 15 + 32} = - \frac{}{17} = - 7 \\ S o l {\begin{matrix} x = - 5 \\ y = - 7 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 13 x - 31 y = - 326 \\ 25 x + 37 y = 146 \end{matrix} \\ x = \frac{| \begin{matrix} - 326 & - 31 \\ 146 & 37 \end{matrix} |}{| \begin{matrix} 13 & - 31 \\ 25 & 37 \end{matrix} |} & = \frac{- 326 (37) - 146 (- 31)}{13 (37) - (25) (- 31)} = \frac{- 12062 + 4526}{481 + 775} = \frac{-}{1256} = - 6 \\ y = \frac{| \begin{matrix} 13 & - 326 \\ 25 & 146 \end{matrix} |}{| \begin{matrix} 13 & - 31 \\ 25 & 37 \end{matrix} |} & = \frac{146 (13) - (- 326) (25)}{13 (37) - (25) (- 31)} = \frac{1898 + 8150}{481 + 775} = \frac{}{1256} = 8 \\ S o l {\begin{matrix} x = - 6 \\ y = 8 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 15 x - 44 y = - 6 \\ 32 y - 27 x = - 1 \end{matrix} \\ {\begin{matrix} 15 x - 44 y = - 6 \\ - 27 x + 32 y = - 1 \end{matrix} \\ x = \frac{| \begin{matrix} - 6 & - 44 \\ - 1 & 32 \end{matrix} |}{| \begin{matrix} 15 & - 44 \\ - 27 & 32 \end{matrix} |} & = \frac{- 6 (32) - (- 1) (- 44)}{15 (32) - (- 27) (- 44)} = \frac{- 192 - 44}{480 - 1188} = \frac{- 236}{-} = \frac{1}{3} \\ y = \frac{| \begin{matrix} 15 & - 6 \\ - 27 & - 1 \end{matrix} |}{| \begin{matrix} 15 & - 44 \\ - 27 & 32 \end{matrix} |} & = \frac{(- 1) (15) - (- 6) (- 27)}{15 (32) - (- 27) (- 44)} = \frac{- 15 - 162}{480 - 1188} = \frac{- 177}{-} = \frac{1}{4} \\ S o l {\begin{matrix} x = \frac{1}{3} \\ y = \frac{1}{4} \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 8 x = - 9 y \\ 2 x + 5 + 3 y = 3 \frac{1}{2} \end{matrix} \\ {\begin{matrix} 8 x + 9 y = 0 \\ 2 x + 3 y = \frac{7}{2} - 5 \end{matrix} \\ {\begin{matrix} 8 x + 9 y = 0 \\ 2 x + 3 y = - \frac{3}{2} & 2 \end{matrix} \\ {\begin{matrix} 8 x + 9 y = 0 \\ 4 x + 6 y = - 3 \end{matrix} \\ x = \frac{| \begin{matrix} 0 & 9 \\ - 3 & 6 \end{matrix} |}{| \begin{matrix} 8 & 9 \\ 4 & 6 \end{matrix} |} & = \frac{6 (0) - 9 (- 3)}{8 (6) - 4 (9)} = \frac{27}{48 - 36} = \frac{}{} = \frac{9}{4} = 2 \frac{1}{4} \\ y = \frac{| \begin{matrix} 8 & 0 \\ 4 & - 3 \end{matrix} |}{| \begin{matrix} 8 & 9 \\ 4 & 6 \end{matrix} |} & = \frac{8 (- 3) - 4 (0)}{8 (6) - 4 (9)} = \frac{- 24}{48 - 36} = \frac{}{12} = 2 \\ S o l {\begin{matrix} x = 2 \frac{1}{4} \\ y = 2 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} a x - b y = - 1 \\ a x + b y = 7 \end{matrix} \\ x = \frac{| \begin{matrix} - 1 & - b \\ 7 & b \end{matrix} |}{| \begin{matrix} a & - b \\ a & b \end{matrix} |} & = \frac{- b - 7 (- b)}{a b - a (- b)} = \frac{- b + 7 b}{a b + a b} = \frac{b}{2 a b} = \frac{3}{a} \\ y = \frac{| \begin{matrix} a & - 1 \\ a & 7 \end{matrix} |}{| \begin{matrix} a & - b \\ a & b \end{matrix} |} & = \frac{7 a - a (- 1)}{a b - a (- b)} = \frac{7 a + a}{a b + a b} = \frac{a}{2 a b} = \frac{4}{b} \\ S o l {\begin{matrix} x = \frac{3}{a} \\ y = \frac{4}{b} \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 3 x - (y + 2) = 2 y + 1 \\ 5 y - (x + 3) = 3 x + 1 \end{matrix} \\ {\begin{matrix} 3 x - y - 2 - 2 y = 1 \\ 5 y - x - 3 - 3 x = 1 \end{matrix} \\ {\begin{matrix} 3 x - 3 y = 1 + 2 \\ 5 y - 4 x = 3 + 1 \end{matrix} \\ {\begin{matrix} 3 x - 3 y = 3 & \frac{1}{3} \\ 5 y - 4 x = 4 \end{matrix} \\ {\begin{matrix} x - y = 1 \\ - 4 x + 5 y = 4 \end{matrix} \\ x = \frac{| \begin{matrix} 1 & - 1 \\ 4 & 5 \end{matrix} |}{| \begin{matrix} 1 & - 1 \\ - 4 & 5 \end{matrix} |} & = \frac{5 (1) - 4 (- 1)}{5 (1) - (- 4) (- 1)} = \frac{5 + 4}{5 - 4} = 9 \\ y = \frac{| \begin{matrix} 1 & 1 \\ - 4 & 4 \end{matrix} |}{| \begin{matrix} 1 & - 1 \\ - 4 & 5 \end{matrix} |} & = \frac{4 (1) - 4 (- 1)}{5 (1) - (- 4) (- 1)} = \frac{4 + 4}{5 - 4} = 8 \\ S o l {\begin{matrix} x = 9 \\ y = 8 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} a x + 2 y = 2 \\ \frac{a x}{2} - 3 y = - 1 & (2) \end{matrix} \\ {\begin{matrix} a x + 2 y = 2 \\ a x - 6 y = - 2 \end{matrix} \\ x = \frac{| \begin{matrix} 2 & 2 \\ - 2 & - 6 \end{matrix} |}{| \begin{matrix} a & 2 \\ a & - 6 \end{matrix} |} & = \frac{2 (- 6) - (- 2) (2)}{a (- 6) - 2 a} = \frac{- 12 + 4}{- 6 a - 2 a} = \frac{- 8}{- 8 a} = \frac{1}{a} \\ y = \frac{| \begin{matrix} a & 2 \\ a & - 2 \end{matrix} |}{| \begin{matrix} a & 2 \\ a & - 6 \end{matrix} |} & = \frac{a (- 2) - 2 a}{a (- 6) - 2 a} = \frac{- 2 a - 2 a}{- 6 a - 2 a} = \frac{- 4 a}{-} = \frac{1}{2} \\ S o l {\begin{matrix} x = \frac{1}{a} \\ y = \frac{1}{2} \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} \frac{x}{4} + \frac{y}{6} = - 4 & (12) \\ \frac{x}{8} - \frac{y}{12} = 0 & (24) \end{matrix} \\ {\begin{matrix} 3 x + 2 y = - 48 \\ 3 x - 2 y = 0 \end{matrix} \\ x = \frac{| \begin{matrix} - 48 & 2 \\ 0 & - 2 \end{matrix} |}{| \begin{matrix} 3 & 2 \\ 3 & - 2 \end{matrix} |} & = \frac{- 2 (- 48) - (2) (0)}{3 (- 2) - 2 (3)} = \frac{96}{- 6 - 6} = \frac{}{- 12} = - 8 \\ y = \frac{| \begin{matrix} 3 & - 48 \\ 3 & 0 \end{matrix} |}{| \begin{matrix} 3 & 2 \\ 3 & - 2 \end{matrix} |} & = \frac{3 (0) - 3 (- 48)}{3 (- 2) - 2 (3)} = \frac{144}{- 6 - 6} = \frac{}{- 12} = - 12 \\ S o l {\begin{matrix} x = - 8 \\ y = - 12 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 3 x + a y = 3 a + 1 \\ \frac{x}{a} + a y = 2 & a \end{matrix} \\ {\begin{matrix} 3 x + a y = 3 a + 1 \\ x + a^{2} y = 2 a \end{matrix} \\ x = \frac{| \begin{matrix} 3 a + 1 & a \\ 2 a & a^{2} \end{matrix} |}{| \begin{matrix} 3 & a \\ 1 & a^{2} \end{matrix} |} & = \frac{a^{2} (3 a + 1) - 2 a (a)}{3 a^{2} - a} = \frac{3 a^{3} + a^{2} - 2 a^{2}}{a (3 a - 1)} = \frac{3 a^{3} - a^{2}}{a (3 a - 1)} = \frac{a^{2} (3 a - 1)}{a (3 a - 1)} = a \\ y = \frac{| \begin{matrix} 3 & 3 a + 1 \\ 1 & 2 a \end{matrix} |}{| \begin{matrix} 3 & a \\ 1 & a^{2} \end{matrix} |} & = \frac{2 a (3) - (3 a + 1)}{3 a^{2} - a} = \frac{6 a - 3 a - 1}{a (3 a - 1)} = \frac{(3 a - 1)}{a (3 a - 1)} = \frac{1}{a} \\ S o l {\begin{matrix} x = a \\ y = \frac{1}{a} \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} \frac{x + 2}{3} - \frac{y - 3}{8} = \frac{5}{6} & (24) \\ \frac{y - 5}{6} - \frac{2 x - 3}{5} = 0 & (30) \end{matrix} \\ {\begin{matrix} 8 (x + 2) - 3 (y - 3) = 20 \\ 5 (y - 5) - 6 (2 x - 3) = 0 \end{matrix} \\ {\begin{matrix} 8 x + 16 - 3 y + 9 = 20 \\ 5 y - 25 - 12 x + 18 = 0 \end{matrix} \\ {\begin{matrix} 8 x - 3 y = 20 - 25 \\ - 12 x + 5 y = 7 \end{matrix} \\ {\begin{matrix} 8 x - 3 y = - 5 \\ - 12 x + 5 y = 7 \end{matrix} \\ x = \frac{| \begin{matrix} - 5 & - 3 \\ 7 & 5 \end{matrix} |}{| \begin{matrix} 8 & - 3 \\ - 12 & 5 \end{matrix} |} & = \frac{- 5 (5) - 7 (- 3)}{8 (5) - (- 12) (- 3)} = \frac{- 25 + 21}{40 - 36} = \frac{- 4}{4} = - 1 \\ y = \frac{| \begin{matrix} 8 & - 5 \\ - 12 & 7 \end{matrix} |}{| \begin{matrix} 8 & - 3 \\ - 12 & 5 \end{matrix} |} & = \frac{8 (7) - (- 12) (- 5)}{8 (5) - (- 12) (- 3)} = \frac{56 - 60}{40 - 36} = \frac{- 4}{4} = - 1 \\ S o l {\begin{matrix} x = - 1 \\ y = - 1 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 3 x - 2 y = 5 \\ m x + 4 y = 2 (m + 1) \end{matrix} \\ x = \frac{| \begin{matrix} 5 & - 2 \\ 2 (m + 1) & 4 \end{matrix} |}{| \begin{matrix} 3 & - 2 \\ m & 4 \end{matrix} |} & = \frac{5 (4) - (- 4) (m + 1)}{3 (4) - (- 2) m} = \frac{20 + 4 m + 4}{12 + 2 m} = \frac{4 m + 24}{2 (m + 6)} = \frac{(m + 6)}{2 (m + 6)} = 2 \\ y = \frac{| \begin{matrix} 3 & 5 \\ m & 2 (m + 1) \end{matrix} |}{| \begin{matrix} 3 & - 2 \\ m & 4 \end{matrix} |} & = \frac{6 (m + 1) - 5 m}{3 (4) - (- 2) m} = \frac{6 m + 6 - 5 m}{12 + 2 m} = \frac{m + 6}{2 (m + 6)} = \frac{1}{2} \\ S o l {\begin{matrix} x = 2 \\ y = \frac{1}{2} \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} 2 x - \frac{2 y + 3}{17} = y + 2 & (17) \\ 3 y - \frac{4 x + 1}{21} = 3 x + 5 & (21) \end{matrix} \\ {\begin{matrix} 34 x - 2 y - 3 = 17 y + 34 \\ 63 y - 4 x - 1 = 63 x + 105 \end{matrix} \\ {\begin{matrix} 34 x - 2 y - 17 y = 3 + 34 \\ 63 y - 4 x - 63 x = 1 + 105 \end{matrix} \\ {\begin{matrix} 34 x - 19 y = 37 \\ - 67 x + 63 y = 106 \end{matrix} \\ x = \frac{| \begin{matrix} 37 & - 19 \\ 106 & 63 \end{matrix} |}{| \begin{matrix} 34 & - 19 \\ - 67 & 63 \end{matrix} |} & = \frac{37 (63) - (- 19) (106)}{34 (63) - (- 19) (- 67)} = \frac{2331 + 2014}{2142 - 1273} = \frac{}{869} = 5 \\ y = \frac{| \begin{matrix} 34 & 37 \\ - 67 & 106 \end{matrix} |}{| \begin{matrix} 34 & - 19 \\ - 67 & 63 \end{matrix} |} & = \frac{34 (106) - (- 67) (37)}{34 (63) - (- 19) (- 67)} = \frac{3604 + 2479}{2142 - 1273} = \frac{}{869} = 7 \\ S o l {\begin{matrix} x = 5 \\ y = 7 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} \frac{x + y}{x - y} = 4 & (x - y) \\ \frac{x - y - 1}{x + y + 1} = \frac{1}{9} & 9 (x + y + 1) \end{matrix} \\ {\begin{matrix} x + y = 4 (x - y) \\ 9 (x - y - 1) = x + y + 1 \end{matrix} \\ {\begin{matrix} x + y = 4 x - 4 y \\ 9 x - 9 y - 9 = x + y + 1 \end{matrix} \\ {\begin{matrix} x + y - 4 x + 4 y = 0 \\ 9 x - 9 y - x - y = 9 + 1 \end{matrix} \\ {\begin{matrix} - 3 x + 5 y = 0 \\ 8 x - 10 y = 10 & \frac{1}{2} \end{matrix} \\ {\begin{matrix} - 3 x + 5 y = 0 \\ 4 x - 5 y = 5 \end{matrix} \\ x = \frac{| \begin{matrix} 0 & 5 \\ 5 & - 5 \end{matrix} |}{| \begin{matrix} - 3 & 5 \\ 4 & - 5 \end{matrix} |} & = \frac{0 (- 5) - 5 (5)}{- 3 (- 5) - 5 (4)} = \frac{- 25}{15 - 20} = \frac{-}{- 5} = 5 \\ y = \frac{| \begin{matrix} - 3 & 0 \\ 4 & 5 \end{matrix} |}{| \begin{matrix} - 3 & 5 \\ 4 & - 5 \end{matrix} |} & = \frac{0 (4) - 3 (5)}{- 3 (- 5) - 5 (4)} = \frac{- 15}{15 - 20} = \frac{-}{- 5} = 3 \\ S o l {\begin{matrix} x = 5 \\ y = 3 \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} x - y = 2 b \\ \frac{x}{a + b} + \frac{y}{a - b} = 2 \end{matrix} \\ {\begin{matrix} x - y = 2 b \\ \frac{x (a - b) + y (a + b)}{(a + b) (a - b)} = 2 \end{matrix} \\ {\begin{matrix} x - y = 2 b \\ a x - b x + a y + b y = 2 (a^{2} - b^{2}) \end{matrix} \\ {\begin{matrix} x - y = 2 b \\ (a - b) x + (a + b) y = 2 (a^{2} - b^{2}) \end{matrix} \\ x = \frac{| \begin{matrix} 2 b & - 1 \\ 2 (a^{2} - b^{2}) & a + b \end{matrix} |}{| \begin{matrix} 1 & - 1 \\ a - b & a + b \end{matrix} |} & = \frac{2 b (a + b) + 2 (a^{2} - b^{2})}{a + b + (a - b)} = \frac{2 (a + b) [b + a - b]}{a + b + a - b} = \frac{2 (a + b) a}{2 a} = a + b \\ y = \frac{| \begin{matrix} 1 & 2 b \\ a - b & 2 (a^{2} - b^{2}) \end{matrix} |}{| \begin{matrix} 1 & - 1 \\ a - b & a + b \end{matrix} |} & = \frac{2 (a^{2} - b^{2}) - 2 b (a - b)}{a + b + (a - b)} = \frac{2 (a - b) [a + b - b]}{a + b + a - b} = \frac{2 (a - b) a}{2 a} = a - b \\ S o l {\begin{matrix} x = a + b \\ y = a - b \end{matrix} \end{matrix}$
$\begin{matrix} {\begin{matrix} \frac{x + 9}{x - 9} = \frac{y + 21}{y + 39} \\ \frac{x + 8}{x - 8} = \frac{y + 19}{y + 11} \end{matrix} \\ {\begin{matrix} (x + 9) (y + 39) = (y + 21) (x - 9) \\ (x + 8) (y + 11) = (y + 19) (x - 8) \end{matrix} \\ {\begin{matrix} x y + 39 x + 9 y + 351 = x y - 9 y + 21 x - 189 \\ x y + 11 x + 8 y + 88 = x y - 8 y + 19 x - 152 \end{matrix} \\ {\begin{matrix} 39 x + 9 y + 9 y - 21 x = - 351 - 189 \\ 11 x + 8 y + 8 y - 19 x = - 88 - 152 \end{matrix} \\ {\begin{matrix} 18 x + 18 y = - 540 & \frac{1}{18} \\ - 8 x + 16 y = - 240 & - \frac{1}{8} \end{matrix} \\ {\begin{matrix} x + y = - 30 \\ x - 2 y = 30 \end{matrix} \\ x = \frac{| \begin{matrix} - 30 & 1 \\ 30 & - 2 \end{matrix} |}{| \begin{matrix} 1 & 1 \\ 1 & - 2 \end{matrix} |} & = \frac{- 30 (- 2) - (30) (1)}{1 (- 2) - (1) (1)} = \frac{60 - 30}{- 2 - 1} = \frac{}{- 3} = - 10 \\ y = \frac{| \begin{matrix} 1 & - 30 \\ 1 & 30 \end{matrix} |}{| \begin{matrix} 1 & 1 \\ 1 & - 2 \end{matrix} |} & = \frac{30 (1) - (- 30) (1)}{1 (- 2) - (1) (1)} = \frac{30 + 30}{- 2 - 1} = \frac{}{- 3} = - 20 \\ S o l {\begin{matrix} x = - 10 \\ y = - 20 \end{matrix} \end{matrix}$