2.the solution set of x +y =5 and x-y =3 x=1, y=5
x=2 ,y=3
x=4 , y=1

3.check the solution x=5 and y=1 is true for x+3y =15 and 2x+7y=19
yes
no

4.find solution of x/2 +y/3 =0 and x+2y=1

x=1 y=3
x=-1/2 ,y=3/4
x=1/2 and y=-3/4

5.x=1/2 and y =x +1 then solution is

1 ,2
1/2 ,1/2
1/2 . 3/2

6.3x +y =5 and x -3y =5 has solution

x=1 ,y= 2
x=2 , y=3
x=2 , y=-1

plz hlp

Check the solution below

Step-by-step explanation:

2) Given the equation

x +y =5... 1 and

x-y =3 ... 2

Add both equations

x+x = 5+3

2x = 8

x = 8/2

x = 4

Substitute x = 4 into 1:

From 1: x+y = 5

4+y= 5

y = 5-4

y = 1

3) Given

x+3y =15 ... 1

2x+7y=19 2

From 2: x = 15-3y

Substitute into 2

2(15-3y)+7y = 19

30-6y+7y = 19

30+y = 19

y = 19-30

y = -11

Substitute y=-11 into x = 15-3y

x =15-3(-11)

x = 15+33

x = 48

The solution set is (48, -11)

4) given

x/2 +y/3 =0 and x+2y=1

From 1

(3x+2y)/6 = 0

3x+2y = 0.. 3

x+2y= 1... 4

From 4: x = 1-2y

Substutute

3(1-2y) +2y = 0

3-6y+2y = 0

3 -4y = 0

4y = 3

y = 3/4

Since x = 1-2y

x = 1-2(3/4)

x = 1-3/2

x= -1/2

The solution set is (-1/2, 3/4)

5) Given

5.x=1/2 and y =x +1 then solution is

We already know the vkue of x

Get y

y= x+1

y = 1/2 + 1

y = 3/2

Hence the solution set is (1/2, 3/2)

6) Given

3x +y =5 and x -3y =5

From 3; x = 5+3y

Substitute into 1;

3(5+3y)+y = 5

15+9y+y = 5

10y = 5-15

10y =-10

y = -1

Get x;

x = 5+3y

x = 5+3(-1)

x = 5-3

x = 2

Hence two solution set is (2,-1)

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5