Answer:What are the equivalence classes of the equivalence relations in Exercise 3? A binary relation defined on a set S is said to be equivalence relation if it is reflexive, symmetric and transitive. An equivalence relation defined on a set S, partition the set into disjoint equivalence classes
Answer:
x = 31/9 and y = 5/3
Step-by-step explanation:
It is given that,
3x - 2y = 7 -----(1)
3x + 4y = 17 ----(2)
<u>To find the solution by elimination method</u>
Step 1: Subtract eq(2) from eq(1)
3x - 2y = 7 -----(1)
<u> 3x + 4y = 17 </u>----(2)
0 - 6y = -10
6y = 10
y = 10/6 = 5/3
Step 2: Substitute the value of y in eq (1)
3x - 2y = 7 -----(1)
3x - 2*(5/3) = 7
3x = 7 + 10/3
3x = 31/3
x = 31/9
Therefore x = 31/9 and y = 5/3
Answer:
m = -6
Step-by-step explanation:
3m= 5(m + 3)-3
Distribute
3m = 5m +15 -3
Combine like terms
3m = 5m +12
Subtract 5m
3m-5m = 5m+12-5m
-2m = 12
Divide by -2
-2m/-2 = 12/-2
m = -6
