# 発展・３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 3test

(1)

(1) −

3 10x+

7

10x= (2) 7 4z +

7

4z = (3) 1 4x−

5 4x=

(4) 5

9y− 2

9y = (5) 11

6 y− 5

6y= (6) 4 9c−

5 9c=

(7) −

1 9x−

7

9x= (8) 8 9c+

4

9c= (9) − 5 4x+

7 4x=

(10) −

2 3y−

7 3 +

5 3y−

5

3 = (11) −

9 10 +

7 10b−

7 10 −

1 10b=

(12) 1

8x+ 3 8x+

1 8 +

7

8 = (13) −

1 10 +

3 10c+

7 10 +

1 10 c=

## 2.

(1) −7z−5

6 +

z−5

6 = (2)

b1 4 +

b+ 5 4 =

(3) 4b5+ 6 + −9b−9

5 = (4)

9x+ 3 10 +

3x+ 7 10 =

(5) 7z−5

6 +

z5

6 = (6)

x2 3 +

5x+ 1 3 =

## 3.

(1) −2c+ 4

9 −

−5c−8

9 = (2)

b−9

5 −

−3b+ 7

5 =

(3) 7b−3

8 −

3b+ 5

8 = (4)

4a4 5 −

3a8 5 =

(5) −7c−7

9 −

−8c−1

9 = (6)

5x+ 3 8 −

(2)

(1) −

3z − 6z (2) 4x 6x

## 5.

(1) x+ 36 + 13x−1

2 = (2)

b19 3 +

−b+ 13

6 =

(3) −5a−1

3 +

−3a+ 9

4 = (4)

−7x+ 7

4 +

−5x−7

6 =

(1) −7y−16

4 −

4y+ 11

5 = (2)

z1 6 −

−13z −5

2 =

(3) −4c+ 3

5 −

−11c+ 1

2 = (4)

−5x+ 13

6 −

(3)

(1) −12c+ 10

7 +

−10c−1

7 = (2)

7b1 10 −

7b9 10 =

(3) 3c+ 7

2 +

5c7

2 = (4)

a7 4 −

−5a−3

4 =

## 8.

(1) −5x+ 1

6 −

−13x−5

2 = (2)

a+ 9 5 −

6a+ 13 4 =

(3) 3c2+ 4 + 3c+ 10

3 = (4)

3x+ 1 4 −

9x3 2 =

(5) 3c−5

4 +

9c+ 1

2 = (6)

−3b+ 11

2 +

−b−11

(4)

(1) −

3 10x+

7 10x=

(2) 7 4z +

7 4z =

(3) 1

4x− 5

4x=

(4) 59y 2

9y =

(5) 11

6 y− 5

6y=

(6) 4 9c−

5

9c=

(7) −

1 9x−

7

9x=

(8) 8 9c+

4 9c=

(9) − 5 4x+

7 4x=

(10) −

2 3y−

7 3 +

5 3y−

5 3 =

(11) −

9 10 +

7 10b−

7 10 −

1 10b=

(12) 1

8x+ 3 8x+

1 8 +

7 8 =

(13) −

1 10 +

3 10c+

7 10 +

1 10 c=

## 2.

(1) −7z−5

6 +

z−5

6 =

(2) b−1

4 +

b+ 5 4 =

### 2

(3) 4b5+ 6 + −9b−9

5 =

(4) 9x+ 3

10 +

3x+ 7 10 =

(5) 7z−5

6 +

z5 6 =

(6) x−2

3 +

5x+ 1 3 =

(1) −2c+ 4

9 −

−5c−8

9 =

(2) b−9

5 −

−3b+ 7

5 =

(3) 7b−3

8 −

3b+ 5

8 = (4)

4a4 5 −

(5)

(1) −

1 3z −

5

6z =

(2) 7 4x+

1 6x=

## 5.

(1) x+ 36 + 13x−1

2 =

(2) b−19

3 +

−b+ 13

6 =

(3) −5a−1

3 +

−3a+ 9

4 =

(4) −7x+ 7

4 +

−5x−7

6 =

(1) −7y−16

4 −

4y+ 11 5 =

(2) z−1

6 −

−13z −5

2 =

(3) −4c+ 3

5 −

−11c+ 1

2 =

(4) −5x+ 13

6 −

9x1 4 =

(6)

(1)

7 7

(2)

10 − 10 =

(3) 3c+ 7

2 +

5c7 2 =

(4) a−7

4 −

−5a−3

4 =

## 8.

(1) −5x+ 1

6 −

−13x−5

2 =

(2) a+ 95

6a+ 13 4 =

### 20

(3) 3c2+ 4 + 3c+ 10 3 =

(4) 3x4+ 1 −

9x3 2 =

(5) 3c−5

4 +

9c+ 1

2 = (6)

−3b+ 11

2 +

−b−11

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