# 発展・２回目 文字式の足し算・引き算 数学・算数の教材公開ページ hatten0103 2test

(1)

## 1.

2y− 7

2y = (2) 11

6 x+ 11

6 x= (3) − 9 10y−

1 10y=

(4) −

3 8a+

7

8a = (5) − 5 2x+

5

2x= (6) − 1 4x+

5 4x=

(7) −

1 6c−

11

6 c= (8) 5 8x+

5

8x= (9) 8 3z −

8 3z =

(10) −

7 3x−

1 3x+

2 3 −

1

3 = (11) − 5 4a−

5 4a−

3 4 − 5 4 = (12) − 3 8y+

5 8y−

5 8 +

1

8 = (13) − 1 5c−

1 5c−

3 5 −

2 5 =

4 +

−x−3

4 = (2)

−11y+ 7

6 +

y−11

6 =

(3) −3z−5

8 +

z+ 7

8 = (4)

−8x+ 2

3 +

−x+ 5

3 =

(5) −12x+ 4

7 +

3x+ 2

7 = (6)

5c1 9 +

7c+ 5 9 =

## 3.

9 −

−2b+ 2

9 = (2)

7x+ 3 8 −

x+ 5 8 =

(3) y−7

4 −

−y−3

4 = (4)

−9c−9

10 −

−7c+ 7

10 =

(5) −x+ 3 − −

7x−3 = (6) a−3 − a

(2)

(1) −

4z 2z (2) −4c 2c

## 5.

3 +

−b−1

6 = (2)

−5a−10

6 +

a+ 10 3 =

(3) c+ 13 + −5c+ 7

6 = (4)

−b−7

6 +

−7b−4

3 =

4 −

y3

2 = (2)

2z+ 13 5 −

−8z+ 1

4 =

(3) a−10

6 −

2a7

3 = (4)

b+ 5 6 −

−7b−5

(3)

3 +

4x+ 2

3 = (2)

−3x−7

10 −

x9 10 =

(3) −7y−2

3 +

−4y−1

3 = (4)

9c3 10 −

7c9 10 =

3 −

5a−7

6 = (2)

5x−1

2 +

−3x−16

5 =

(3) −x+ 1

6 +

x+ 4

3 = (4)

−5y−7

3 +

−5y−1

6 =

(5) −3c−1

4 −

5c−1

6 = (6)

5a+ 1 3 −

−5a+ 1

(4)

(1) 3 2y−

7

2y =

(2) 11 6 x+

11 6 x=

(3) −

9 10y−

1

10y=

(4) −

3 8a+

7 8a =

(5) −

5 2x+

5

2x=

(6) − 1 4x+

5

4x=

(7) −

1 6c−

11

6 c=

(8) 5 8x+

5 8x=

(9) 83z −

8

3z =

(10) −

7 3x−

1 3x+

2 3 − 1 3 =

(11) −

5 4a−

5 4a−

3 4 − 5 4 =

(12) −

3 8y+

5 8y−

5 8 +

1 8 =

(13) −

1 5c−

1 5c−

3 5 − 2 5 =

4 +

−x−3

4 =

(2) −11y+ 7

6 +

y−11

6 =

(3) −3z−5

8 +

z+ 7 8 =

(4) −8x+ 2

3 +

−x+ 5

3 =

(5) −12x+ 4

7 +

3x+ 2 7 =

(6) 5c−1

9 +

7c+ 5 9 =

9 −

−2b+ 2

9 =

(2) 7x8+ 3 − x

+ 5 8 =

(3) y−7

4 −

−y−3

4 = (4)

−9c−9

10 −

−7c+ 7

(5)

(1) −

1 4z +

5 2z =

(2) −

3 4c+

5 2c=

## 5.

3 +

−b−1

6 =

(2) −5a−10

6 +

a+ 10 3 =

### 6

(3) c+ 13 + −5c+ 7

6 =

(4) −b−7

6 +

−7b−4

3 =

4 −

y3 2 =

(2) 2z+ 135 − −

8z+ 1 4 =

(3) a−10

6 −

2a7 3 =

(4) b+ 5 6 −

−7b−5

2 =

(6)

(1)

3 3

(2)

10 − 10

(3) −7y−2

3 +

−4y−1

3 =

(4) 9c−3

10 −

7c9 10 =

3 −

5a−7

6 =

(2) 5x−1

2 +

−3x−16

5 =

(3) −x+ 1

6 +

x+ 4 3 =

(4) −5y−7

3 +

−5y−1

6 =

(5) −3c−1

4 −

5c−1

6 = (6)

5a+ 1 3 −

−5a+ 1

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