# 2ページ目 文字式の足し算・引き算 数学・算数の教材公開ページ hatten02 1 5test

10

(1)

(1) 1 6x−

4

3x= (2) −

1 6z−

7 4z =

(1) −4a−13 3 +

−5a+ 19

6 = (2)

−3x+ 1 2 +

5x−7 6 =

(3) −5x−1 2 +

−5x−7

6 = (4)

5b+ 9 3 +

3b+ 5 4 =

(1) 5c6+ 3 −

7c+ 1

2 = (2)

−x+ 10 6 −

−8x+ 7 3 =

(3) −2z−7 3 −

−5z −7

6 = (4)

−z−19 2 −

(2)

(1) 16x 4

3x=

(2) −

1 6z−

7

4z =

## 2.

(1) −4a−13 3 +

−5a+ 19 6 =

(2) −3x+ 1 2 +

5x−7 6 =

(3) −5x−1 2 +

−5x−7 6 =

### 3

(4) 5b3+ 9 + 3b+ 5 4 =

(1) 5c6+ 3 −

7c+ 1 2 =

(2) −x+ 10 6 −

−8x+ 7 3 =

(3)

(1) 5 6b−

2

3b= (2) −

5 3x+

5 6x=

## 2.

(1) 2x3+ 7 + x+ 1

6 = (2)

−2y−11 3 +

2y+ 1 5 =

(3) −x+ 3 6 +

−13x+ 7

2 = (4)

7a−19 4 +

−5a−19 6 =

(1) −4x−16 3 −

x11

5 = (2)

5y1 2 −

3y+ 9 4 =

(3) −x−7 3 −

x+ 1

6 = (4)

−5x−5 2 −

(4)

(1) 56b 2 3b=

(2) −

5 3x+

5

6x=

## 2.

(1) 2x3+ 7 + x+ 1 6 =

(2) −2y−11 3 +

2y+ 1 5 =

(3) −x+ 3 6 +

−13x+ 7 2 =

(4) 7a−19 4 +

−5a−19 6 =

(1) −4x−16 3 −

x11 5 =

(2) 5y−1 2 −

3y+ 9 4 =

(5)

(1) 5 6x+

4

3x= (2)

5 6y+

5 3y=

(1) −y+ 3 4 +

−y+ 3

2 = (2)

−c+ 16 4 +

−2c−16 5 =

(3) −7x+ 7 4 +

5x+ 1

2 = (4)

c−4 6 +

7c+ 1 3 =

(1) −c−19 3 −

−5c+ 1

6 = (2)

−c+ 7 6 −

−8c+ 10 3 =

(3) −5x+ 3 6 −

13x5

2 = (4)

−x+ 13 2 −

(6)

(1) 56x+ 4 3x=

(2)

5 6y+

5 3y=

(1) −y+ 3 4 +

−y+ 3 2 =

(2) −c+ 16 4 +

−2c−16 5 =

(3) −7x+ 7 4 +

5x+ 1 2 =

(4) c−4 6 +

7c+ 1 3 =

(1) −c−19 3 −

−5c+ 1 6 =

(2) −c+ 7 6 −

−8c+ 10 3 =

(7)

(1) 1 2y−

3

4y = (2) −

5 2x+

1 4x=

(1) 5b−4 6 +

8b+ 7

3 = (2)

x−13 2 +

−x+ 1 4 =

(3) a−13 2 +

−3a−13

4 = (4)

2z−16 3 +

−3z+ 1 5 =

## 3.

(1) a+ 76 − − 2a1

3 = (2)

a1 4 −

−a+ 1 6 =

(3) −3c−7 2 −

−5c−13

6 = (4)

x+ 13 3 −

(8)

(1) 12y 3

4y =

(2) −

5 2x+

1

4x=

(1) 5b−4 6 +

8b+ 7 3 =

(2) x−13 2 +

−x+ 1 4 =

(3) a−13 2 +

−3a−13 4 =

(4) 2z−16 3 +

−3z+ 1 5 =

## 3.

(1) a+ 76 − − 2a1

3 =

(2) a−1 4 −

−a+ 1 6 =

(9)

(1) − 3 2x−

1

6x= (2)

1 3c+

3 4c=

(1) −x+ 1 2 +

x+ 13

4 = (2)

−4b−9 5 +

−b+ 9 4 =

(3) −5x+ 4 4 +

6x−1

3 = (4)

5a+ 5 4 +

9a+ 3 2 =

(1) 5b+ 1 6 −

−13b−1

2 = (2)

−5b−13 3 −

b13 6 =

(3) −5x+ 1 6 −

x+ 4

3 = (4)

x+ 4 2 −

(10)

(1) − 3 2x−

1

6x=

(2)

1 3c+

3 4c=

## 2.

(1) −x+ 1 2 +

x+ 13 4 =

(2) −4b−9 5 +

−b+ 9 4 =

(3) −5x+ 4 4 +

6x−1 3 =

### 12

(4) 5a4+ 5 + 9a+ 3 2 =

(1) 5b+ 1 6 −

−13b−1 2 =

(2) −5b−13 3 −

b13 6 =

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