Answer: (a^2bc)^2 for the first one
Answer:
(2, 5 )
Step-by-step explanation:
Given the 2 equations
2x + 3y = 19 → (1)
6x + 2y = 22 → (2)
Multiplying (1) by - 3 and adding to (2) will eliminate the x- term
- 6x - 9y = - 57 → (3)
Add (2) and (3) term by term to eliminate x
0 - 7y = - 35
- 7y = - 35 ( divide both sides by - 7 )
y = 5
Substitute y = 5 into either of the 2 equations and solve for x
Substituting into (1)
2x + 3(5) = 19
2x + 15 = 19 ( subtract 15 from both sides )
2x = 4 ( divide both sides by 2 )
x = 2
solution is (2, 5 )
1) it must be fulfilled that the denominator is different from zero, then for this expression
x^2 -9 ⇒ x^2-3^2 = (x-3)(x+3)⇒ (x-3)(x+3) = 0 ⇒ x = 3 o x = -3 , the dominium is all numbers reals except 3 and -3
