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1. (教材练习变式)解方程组$\begin{cases}2x + y = 3①,\\2x - 3y = 4②\end{cases}$时,由① - ②可得( )
A.$-2y = -1$
B.$-2y = 1$
C.$4y = 1$
D.$4y = -1$
A.$-2y = -1$
B.$-2y = 1$
C.$4y = 1$
D.$4y = -1$
答案
1. D
解析
由①-②得：$(2x + y)-(2x - 3y)=3 - 4$，
去括号得：$2x + y - 2x + 3y=-1$，
合并同类项得：$4y=-1$。
D
去括号得：$2x + y - 2x + 3y=-1$，
合并同类项得：$4y=-1$。
D
2. 在解二元一次方程组$\begin{cases}6x + \oplus y = 9①,\\2x + \otimes y = -6②\end{cases}$时,若① - ②可直接消去未知数$y$,则$\oplus$和$\otimes$( )
A.互为倒数
B.大小相等
C.都等于 0
D.互为相反数
A.互为倒数
B.大小相等
C.都等于 0
D.互为相反数
答案
2. B 解析：通过两个方程作减法来消去某个未知数，一定是这两个方程中相同未知数的系数相等.
3. 用加减消元法解方程组$\begin{cases}5x - 2y = 3①,\\x + 2y = -19②.\end{cases}$下列做法正确的是( )
A.① + ②
B.① - ②
C.① + ②×5
D.①×5 - ②
A.① + ②
B.① - ②
C.① + ②×5
D.①×5 - ②
答案
3. A
4. 已知$x$、$y$满足的方程组是$\begin{cases}x + 2y = 2,\\2x + 3y = 7,\end{cases}$则$x + y$的值为 ______ .
答案
4.5解析：②①，得。
解析
$\begin{cases} x + 2y = 2 \quad \textcircled{1} \\ 2x + 3y = 7 \quad \textcircled{2} \end{cases}$
$\textcircled{2} - \textcircled{1}$，得：
$(2x + 3y) - (x + 2y) = 7 - 2$
$2x + 3y - x - 2y = 5$
$x + y = 5$
5
$\textcircled{2} - \textcircled{1}$，得：
$(2x + 3y) - (x + 2y) = 7 - 2$
$2x + 3y - x - 2y = 5$
$x + y = 5$
5
5. 如果$4x^{a + 2b - 5} - 2y^{3a - b - 3} = 8$是二元一次方程,那么$a - b =$
0
.答案
5. 0 解析：由题意，得$\begin{cases}a + 2b - 5 = 1,\\3a - b - 3 = 1,\end{cases}$即$\begin{cases}a + 2b = 6①,\\3a - b = 4②,\end{cases}$②$×2$，得$6a - 2b = 8$③，①$+$③，得$7a = 14$，解得$a = 2$。把$a = 2$代入②，得$3×2 - b = 4$，解得$b = 2$，$\therefore a - b = 2 - 2 = 0$。
解析
由题意，得$\{\begin{array}{l}a + 2b - 5 = 1\\3a - b - 3 = 1\end{array} $，即$\{\begin{array}{l}a + 2b = 6\quad①\\3a - b = 4\quad②\end{array} $
$②×2$，得$6a - 2b = 8\quad③$
$① + ③$，得$7a = 14$，解得$a = 2$
把$a = 2$代入$②$，得$3×2 - b = 4$，解得$b = 2$
$\therefore a - b = 2 - 2 = 0$
0
$②×2$，得$6a - 2b = 8\quad③$
$① + ③$，得$7a = 14$，解得$a = 2$
把$a = 2$代入$②$，得$3×2 - b = 4$，解得$b = 2$
$\therefore a - b = 2 - 2 = 0$
0
6. 已知$\begin{cases}4x - y = 1,\\-x + 4y = 4,\end{cases}$则$(x + y)(x - y)$的值为 ______ .
答案
6. $-1$ 解析
$\begin{cases}4x - y = 1①, \\-x + 4y = 4②.\end{cases}$
$① + ②$，得$3x + 3y = 5$，$\therefore x + y = \dfrac{5}{3}$；
$① - ②$，得$5x - 5y = - 3$，$\therefore x - y = - \dfrac{3}{5}$。
$\therefore (x + y)(x - y) = \dfrac{5}{3}×(-\dfrac{3}{5}) = - 1$。
$\begin{cases}4x - y = 1①, \\-x + 4y = 4②.\end{cases}$
$① + ②$，得$3x + 3y = 5$，$\therefore x + y = \dfrac{5}{3}$；
$① - ②$，得$5x - 5y = - 3$，$\therefore x - y = - \dfrac{3}{5}$。
$\therefore (x + y)(x - y) = \dfrac{5}{3}×(-\dfrac{3}{5}) = - 1$。
7. 用加减法解下列方程组:
(1)$\begin{cases}x + y = 11,\\2x - y = 7;\end{cases}$
(2)$\begin{cases}3x - 2y = 5,\\x + 3y = 9;\end{cases}$
(3)$\begin{cases}x - 3y - 1 = 0,\\4x - 5y - 18 = 0;\end{cases}$
(4)$\begin{cases}4x - 3y = 11,\\2x + y = 13;\end{cases}$
(5)$\begin{cases}3x + 4y = 16,\\5x - 6y = 33;\end{cases}$
(6)$\begin{cases}\dfrac{x}{2} - \dfrac{y + 1}{3} = 1,\\3x + 2y = 40.\end{cases}$
(1)$\begin{cases}x + y = 11,\\2x - y = 7;\end{cases}$
(2)$\begin{cases}3x - 2y = 5,\\x + 3y = 9;\end{cases}$
(3)$\begin{cases}x - 3y - 1 = 0,\\4x - 5y - 18 = 0;\end{cases}$
(4)$\begin{cases}4x - 3y = 11,\\2x + y = 13;\end{cases}$
(5)$\begin{cases}3x + 4y = 16,\\5x - 6y = 33;\end{cases}$
(6)$\begin{cases}\dfrac{x}{2} - \dfrac{y + 1}{3} = 1,\\3x + 2y = 40.\end{cases}$
答案
(1)$\begin{cases}x + y = 11,①\\2x - y = 7;②\end{cases}$
①+②得：$3x=18$，解得$x=6$，
将$x=6$代入①得：$6 + y=11$，解得$y=5$，
$\therefore\begin{cases}x=6\\y=5\end{cases}$
(2)$\begin{cases}3x - 2y = 5,①\\x + 3y = 9;②\end{cases}$
②×3得：$3x + 9y=27$，③
③-①得：$11y=22$，解得$y=2$，
将$y=2$代入②得：$x + 6=9$，解得$x=3$，
$\therefore\begin{cases}x=3\\y=2\end{cases}$
(3)$\begin{cases}x - 3y = 1,①\\4x - 5y = 18;②\end{cases}$
①×4得：$4x - 12y=4$，③
②-③得：$7y=14$，解得$y=2$，
将$y=2$代入①得：$x - 6=1$，解得$x=7$，
$\therefore\begin{cases}x=7\\y=2\end{cases}$
(4)$\begin{cases}4x - 3y = 11,①\\2x + y = 13;②\end{cases}$
②×3得：$6x + 3y=39$，③
①+③得：$10x=50$，解得$x=5$，
将$x=5$代入②得：$10 + y=13$，解得$y=3$，
$\therefore\begin{cases}x=5\\y=3\end{cases}$
(5)$\begin{cases}3x + 4y = 16,①\\5x - 6y = 33;②\end{cases}$
①×3得：$9x + 12y=48$，③
②×2得：$10x - 12y=66$，④
③+④得：$19x=114$，解得$x=6$，
将$x=6$代入①得：$18 + 4y=16$，解得$y=-\dfrac{1}{2}$，
$\therefore\begin{cases}x=6\\y=-\dfrac{1}{2}\end{cases}$
(6)$\begin{cases}3x - 2y = 8,①\\3x + 2y = 40;②\end{cases}$
①+②得：$6x=48$，解得$x=8$，
将$x=8$代入②得：$24 + 2y=40$，解得$y=8$，
$\therefore\begin{cases}x=8\\y=8\end{cases}$
①+②得：$3x=18$，解得$x=6$，
将$x=6$代入①得：$6 + y=11$，解得$y=5$，
$\therefore\begin{cases}x=6\\y=5\end{cases}$
(2)$\begin{cases}3x - 2y = 5,①\\x + 3y = 9;②\end{cases}$
②×3得：$3x + 9y=27$，③
③-①得：$11y=22$，解得$y=2$，
将$y=2$代入②得：$x + 6=9$，解得$x=3$，
$\therefore\begin{cases}x=3\\y=2\end{cases}$
(3)$\begin{cases}x - 3y = 1,①\\4x - 5y = 18;②\end{cases}$
①×4得：$4x - 12y=4$，③
②-③得：$7y=14$，解得$y=2$，
将$y=2$代入①得：$x - 6=1$，解得$x=7$，
$\therefore\begin{cases}x=7\\y=2\end{cases}$
(4)$\begin{cases}4x - 3y = 11,①\\2x + y = 13;②\end{cases}$
②×3得：$6x + 3y=39$，③
①+③得：$10x=50$，解得$x=5$，
将$x=5$代入②得：$10 + y=13$，解得$y=3$，
$\therefore\begin{cases}x=5\\y=3\end{cases}$
(5)$\begin{cases}3x + 4y = 16,①\\5x - 6y = 33;②\end{cases}$
①×3得：$9x + 12y=48$，③
②×2得：$10x - 12y=66$，④
③+④得：$19x=114$，解得$x=6$，
将$x=6$代入①得：$18 + 4y=16$，解得$y=-\dfrac{1}{2}$，
$\therefore\begin{cases}x=6\\y=-\dfrac{1}{2}\end{cases}$
(6)$\begin{cases}3x - 2y = 8,①\\3x + 2y = 40;②\end{cases}$
①+②得：$6x=48$，解得$x=8$，
将$x=8$代入②得：$24 + 2y=40$，解得$y=8$，
$\therefore\begin{cases}x=8\\y=8\end{cases}$
