4. Substitute x=2 into x+3k=-2:

2+3k=-2

3k=-2-2

3k=-4

k=-4/3

5. small. This can avoid scores with large denominator.

6.{x= 1，y=3/2

{x=0，y=3

{x=2，y=0

8.D

The correct answer should be c times the given equation, and c is an arbitrary constant.

9.(3){y+2= 1-x，① 3x=2y=-3 ②

Solution: From ①, get

y=- 1-x ③

Substitute ③ into ② to get.

(2) What is it? I regard ② as 3x+2y=-3. )

3x+2(- 1-x)=-3

3x-2-2x=-3

x-2=-3

x=-3+2

x=- 1

Substitute x=- 1 into ③ to get.

y=- 1-(- 1)=0

So the solution of this equation group is

{x=- 1 y=0

(5){3m+2n=5m+2，① 2(3m+2n)= 1 1m+7 ②

Solution: From ①, get

m=n- 1 ③

Substitute ③ into ② to get.

2[3(n- 1)+2n]= 1 1(n- 1)+7

2(3n-3+2n)= 1 1n- 1 1+7

2(5n-3)= 1 1n-4

10n-6= 1 1n-4

10n- 1 1n=-4+6

-n=2

n=-2

Substitute n=-2 into ③ to get.

m=-2- 1=-3

So the solution of this equation group is

{m=-3 n=-2

10.{x-3y=2，① x:y=4:3 ②

Solution: From ①, get

x=3y+2 ③

Substitute ③ into ② to get.

(3y+2):y=4:3

3y+2=4y/3

3y-4y/3=-2

5y/3=-2

y=-6/5

Substitute y=-6/5 into ③ to get.

x=3(-6/5)+2=-8/5

So the solution of this equation group is

{x=-8/5 y=-6/5

12.A

Substitute {x=2 y= 1 into {ax+by=7 ax-by= 1.

{2a+b=7，① 2a-b= 1 ②

From (1), we get

b=7-2a ③

Substitute ③ into ② to get.

2a-(7-2a)= 1

2a-7+2a= 1

4a-7= 1

4a= 1+7

4a=8

a=2

Substitute a=2 into ③ and you get

b=7-2X2=7-4=3

a-b=2-3=- 1

15. Suppose A and B have X and Y books respectively, then

{x=2y-30，① x- 10=y+ 10 ②

Substitute ① into ② to get.

2y-30- 10=y+ 10

2y-40=y+ 10

2y-y= 10+40

y=50

Substitute y=50 into ① to obtain

x=2x50-30=70

So it turns out that A and B each have seventy or fifty books.

16.p=-4/3，q=8/3

Substituting {x=-3 y=3 into {px-y= 1 2x+qy=2, we get.

{-3p-3= 1，① 2x(-3)+3q=2 ②

From (1), we get

p=-4/3

From 2, get

q=8/3

17.{-4=-2k+b，① -3=0k+b ②

From 2, get

b=-3 ③

Substitute ③ into ① to get.

-4=-2k-3

2k=-3+4

2k= 1

k= 1/2

So y=kx+b is y=x/2-3.

(1) when x=2, y = 2/2-3 =1-3 =-2;

(3) When y=2,

2=x/2-3

-x/2=-3-2

-x/2=-5

x= 10

18.{m/2+n/3= 13，① m/3-n/4=3 ②

From (1), we get

m=26-2n/3 ③

Substitute ③ into ① to get.

(26-2n/3)/3-n/4=3

26/3-2n/9-n/4=3

26/3- 17n/36=3

- 17n/36=3-26/3

- 17n/36=- 17/3

n= 12

Substitute n= 12 into ③ to get.

m = 26-2x 12/3 = 26-8 = 18

So the solution of this equation group is

{m= 18 n= 12

20.{2x+y=28，① y=2x ②

Substitute ② into ① to get.

2x+2x=28

4x=28

x=7

Substitute x=7 into ②, and you get

y=2x7= 14

So {x+y=5k x-y=-5k/2+b is.

{7+ 14=5k，③ 7- 14=-5k/2+b ④

Obtain from ③

k=2 1/5 ⑤

Substitute ⑤ into ⑤ to get it.

-7=-5x(2 1/5)/2+b

-7=-2 1/2+b

-b=-2 1/2+7

-b=-7/2

b=7/2

∴k=2 1/5,b=7/2

20. We walk X kilometers and ride Y kilometers respectively.

{x+y=28，① x/4+y/36= 1 ②

From (1), we get

x=28-y ③

Substitute ③ into ② to get.

(28-y)/4+y/36= 1

7-y/4+y/36= 1

7-2y/9= 1

-2y/9= 1-7

-2y/9=-6

y=27

Substitute y=27 into ③:

x=28-27= 1

So walking and cycling are 1 and 27 kilometers respectively.

27. There are X and Y medium and small cars.

{x+y=50，① 6x+4y=230 ②

By 1:

x=50-y ③

Substitute ③ into ② to get.

6(50-y)+4y=230

300-6y+4y=230

300-2y=230

-2y=230-300

-2y=-70

y=35

Substitute y=35 into ③ to get.

x=50-35= 15

So there are 15 and 35 small and medium-sized cars respectively.