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Problem 1 :  Option B, x = 9

Problem 2:  Option A, r = 45

Problem 3:  Option D, m = 7

Problem 4:  Option B, b/2 - 4 = 9; Total Bill was $26

Problem 5:  Option C, 3h + 5 = 11; h = 2

Step-by-step explanation:

Problem 1

Step 1:  Add 3 to both sides

2x - 3 + 3 = 15 + 3

2x = 18

Step 2:  Divide both sides by 2

2x / 2 = 18 / 2

x = 9

 Option B, x = 9

Problem 2

Step 1:  Add 4 to both sides

r/3 - 4 + 4 = 11 + 4

r/3 = 15

Step 2:  Multiply both sides by 3

r/3 * 3 = 15 * 3

r = 45

 Option A, r = 45

Problem 3

Step 1:  Subtract 7 from both sides

7 + 3m - 7 = 28 - 7

3m = 21

Step 2:  Divide both sides by 3

3m / 3 = 21 / 3

m = 7

 Option D, m = 7

Problem 4

Step 1:  Make an expression

b/2 -> You and your friend

-4 -> You owe to your friend

9 -> Total your friend gives you

Equation:  b/2 - 4 = 9

Step 2:  Add 4 to both sides

b/2 - 4 + 4 = 9 + 4

b/2 = 13

Step 3:  Multiply both sides by 2

b/2 * 2 = 13 * 2

b = $26

 Option B, b/2 - 4 = 9; Total Bill was $26

Problem 5

Step 1:  Make an expression

3h -> 3 dollars an hour

+5 -> Initial cost

11 -> Total Cost

Equation:  3h + 5 = 11

Step 2:  Subtract 5 from both sides

3h + 5 - 5 = 11 - 5

3h = 6

Step 3:  Divide both sides by 3

3h / 3 = 6 / 3

h = 2

 Option C, 3h + 5 = 11; h = 2

Step-by-step explanation:

1.

2x - 3 = 15

2x - 3 + 3 = 15 + 3

2x = 18

x = 18/2

x = 9

2 .  r/3 - 4 = 11

r/3 - 4 + 4 = 11 + 4

r/3 = 15

r = 15 * 3

r = 45

3 .  7 + 3m = 28

7 + 3m - 7 = 28 - 7

3m = 21

m = 21/3

m = 7

4 .   b/2 - 4 = 9

let b/2  You and your friend

-4  You owe to your friend

9  Total your friend gives you

b/2 - 4 + 4 = 9 + 4

b/2 = 13

b = 13 * 2

b = $26

5 .  3h + 5 = 11

let

3h = 3 dollars per hour

+5  Initial cost

11  Total Cost

3h + 5 - 5 = 11 - 5

3h = 6

h = 6 / 3

h = 2

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)