The algebra of triangles 1-5
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See below.

Step-by-step explanation:

1.

Corresponding angles of similar triangles are congruent.

m<C = m<C'

4x - 30 = x + 15

3x = 45

x = 15

m<C = 4x - 30 = 4(15) - 30 = 60 - 30 = 30

m<C' = m<C = 30

m<A = 2x + 35 = 2(15) + 35 = 30 + 35 = 65

2.

Method: SAS

BC = AD

6a + 20 = a + 60

5a = 40

a = 8

AB = 5a - 12 = 5(8) - 12 = 40 - 12 = 28

BC = 6a + 20 = 6(8) + 20 = 48 + 20 = 68

CD = AB = 28

AD = a + 60 = 8 + 60 = 68

3.

Method: SAS

2x + 3y = 34

4x - y = 40

Multiply the second equation by 3:

12x - 3y = 120

Add this new equation to the first original equation to eliminate y.

14x = 154

x = 11

Substitute 11 for x in the second original equation and solve for y.

4x - y = 40

4(11) - y = 40

44 - y = 40

-y = -4

y = 4

4.

Method: ASA

2d + 10 = 45 - 3d

5d = 35

d = 7

AC = 45 - 3d = 45 - 3(7) = 45 - 21 = 24

EC = 2d + 10 = 2(7) + 10 = 14 + 10 = 24

DE = d + 8 = 7 + 8 = 15

5.

2x - 59 = x + 31

x = 90

DO = x + 31 = 90 + 31 = 121

OG = 2x - 59 = 2(90) - 59 = 180 - 59 = 121

DG = x - 7 = 90 - 7 = 83

answer: can you show the options ?

step-by-step explanation: