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12
解:​$\begin {cases}2x - 5y = 6①\\2x - 8y + 9 = 3②\end {cases}$​
把①代入②,得
​$6 + 5y - 8y + 9 = 3,$​
整理得​$-3y = -12,$​
解得​$y = 4,$​
把​$y = 4$​代入​$①,$​得
​$2x - 5×4 = 6,$​
解得​$x = 13,$​
所以原方程组的解是
​$\begin {cases}x = 13,\\y = 4.\end {cases}$​
解:
​$\begin {cases}\dfrac {x}{3}=\dfrac {y}{5}① \\3x + 4y = 58②\end {cases}$​
由①,得​$x = \frac {3}{5}y. ③$​
把③代入②,得
​$\frac {9}{5}y + 4y = 58,$​
解得​$y = 10. $​
把​$y = 10$​代入​$③,$​得
​$x = 6. $​
所以原方程组的解是
​$\begin {cases}x = 6, \\y = 10.\end {cases}$​
解:方程组整理,得​$\begin {cases}-x + 7y = 4 ① \\2x + y = 3 ②\end {cases}$​
由①,得​$x = 7y - 4. ③$​
把③代入②,得​$2(7y - 4) + y = 3,$​
解得​$y=\frac {11}{15}.$​
把​$y=\frac {11}{15}$​代入​$③,$​得​$x=\frac {17}{15}.$​
所以原方程组的解是​$\begin {cases}x=\frac {17}{15}, \\y =\frac {11}{15}.\end {cases}$​
解:设最初报名时女生有​$x$​人,男生有​$y$​人,
依题意得​$\begin {cases}4x = 3y\\x + 15 = 2y\end {cases},$​解得​$\begin {cases}x = 9\\y = 12\end {cases}。$​
答:最初报名时男生有​$12$​人,女生有​$9$​人。
C
解:​$ (1) $​把方程​$②$​变形为​$3(3x - 2y)+2y = 19 ③,$​
把①代入③,得​$15 + 2y = 19,$​
所以​$y = 2,$​
把​$y = 2$​代入​$①,$​得​$x = 3,$​
则方程组的解为​$\begin {cases}x = 3 \\y = 2\end {cases}。$​
​$ (2) $​由​$3x^2-2xy + 12y^2=47,$​得
​$3(x^2+4y^2) = 47 + 2xy,$​
即​$x^2+4y^2=\frac {47 + 2xy}{3},$​
由​$2x^2+xy + 8y^2=36,$​得
​$2x^2+8y^2=36 - xy,$​
则​$2×\frac {47 + 2xy}{3}=36 - xy,$​解得​$xy = 2,$​
则​$x^2+4y^2=17。$​