Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1324

а) $$ \left\{\begin{array} { l } { ( 2 x - y ) ( 2 x + y ) = 3 } \\ { 2 y - 3 ( x + y ) = - 4 } \end{array} \left\{\begin{array}{l} 4 x^2-y^2=3 \\ 2 y-3 x-3 y=-4 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { 4 x ^ { 2 } - y ^ { 2 } = 3 } \\ { y = 4 - 3 x } \end{array} \left\{\begin{array}{l} 4 x^2-(4-3 x)^2=3 \\ y=4-3 x \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { 4 x ^ { 2 } - 1 6 + 2 4 x - 9 x ^ { 2 } - 3 = 0 } \\ { y = 4 - 3 x } \end{array} \left\{\begin{array}{l} -5 x^2+24 x-19=0 \\ y=4-3 x \end{array}\right.\right. $$
    −5x² + 24x − 19 = 0     D = b² − 4ac = 576 − 4 · (−5) · (−19) = 576 − 380 = 196
    x₁ = $$ \frac{-b+\sqrt{D}}{2 a} $$ = −24 + 14/−10 = −10/−10 = 1
    x₂ = $$ \frac{-b-\sqrt{D}}{2 a}= $$ = −24 − 14/−10 = −38/−10 = 3,8
    $$ \left\{\begin{array} { l } { x _ { 1 } = 1 } \\ { y _ { 1 } = 4 - 3 · 1 } \end{array} \left\{\begin{array}{l} x_1=1 \\ y_1=1 \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { x _ { 1 } = 3 , 8 } \\ { y _ { 1 } = 4 - 3 · 3 , 8 } \end{array} \left\{\begin{array}{l} x_1=3,8 \\ y_1=-7,4 \end{array}\right.\right. $$
    (1; 1), (3,8; −7,4)
б) $$ \left\{\begin{array} { l } { 2 ( x - y ) + y = 5 } \\ { ( 2 x - y ) ^ { 2 } = 5 x + 1 5 } \end{array} \left\{\begin{array} { l } { 2 x - 2 y + y = 5 } \\ { ( 2 x - y ) ^ { 2 } = 5 x + 1 5 } \end{array} \left\{\begin{array}{l} y=2 x-5 \\ (2 x-(2 x-5))^2=5 x+15 \end{array}\right.\right.\right. $$
    $$ \left\{\begin{array} { l } { y = 2 x - 5 } \\ { ( 2 x - 2 x + 5 ) ^ { 2 } = 5 x + 1 5 } \end{array} \left\{\begin{array}{l} y=2 x-5 \\ 5^2=5 x+15 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { y = 2 x - 5 } \\ { 5 x = 2 5 - 1 5 } \end{array} \left\{\begin{array} { l } { y = 2 x - 5 } \\ { 5 x = 1 0 } \end{array} \left\{\begin{array} { l } { y = 2 } \\ { x = 2 } \end{array} \left\{\begin{array}{l} y=-1 \\ x=2 \end{array}\right.\right.\right.\right. $$
    (2; −1)