10x+y-3(x+y)=23 ( 1)

5(x+y)+ 1= 10x+y (2)

From (1)(2)

10x+y-3x-3y=23 7x-2y=23 (3)

5x+5y+ 1- 10x-y = 0 5x-4y = 1(4)

(3)× 2-( 1)

9x=45

x=5

Substitute x=5 into (3).

7×5-2y=23

-2y=- 12

y=6

A: This number is 56.

2. Solution: Let the original two addends be x and y respectively.

10x+y=242 ( 1)

x+ 10y=34 1 (2)

( 1)× 10-(2)

99x=2079

x=2 1

Substitute x=2 1 into (2).

2 1+ 10y=34 1

10y=320

y=32

A: The original two addends were 2 1 and 32 respectively.

3. Solution: Let Xiaoying go uphill and downhill with X and Y points respectively.

4.8km/h = 80m/min12km/h = 200m/min.

x+y= 16 ( 1)

80x+200y= 1880 (2)

( 1)×200-(2)

120x= 1320

x= 1 1

Substitute x= 1 1 (1).

1 1+y= 16

y=5

Answer: Xiaoying uses 1 1 for uphill and 5 for downhill.

4. Solution: Suppose you need two kinds of sweets, X and Y kilograms respectively.

x+y= 100 ( 1)

18x+ 10y = 15× 100(2)

( 1)× 18-(2)

8y=300

y=37.5

Substitute y=37.5 (1).

x+37.5= 100

x=62.5

Answer: You need 62.5 kilograms and 37.5 kilograms of candy.