Answer:
(- 5, 1 )
Step-by-step explanation:
- 6x - 14y = 16 → (1)
- 2x + 7y = 17 → (2)
Multiplying (2) by - 3 and adding to (1) will eliminate the x- term
6x - 21y = - 51 → (3)
Add (1) and (3) term by term to eliminate x
0 - 35y = - 35
- 35y = - 35 ( divide both sides by - 35 )
y = 1
Substitute y = 1 into either of the 2 equations and solve for x
Substituting into (1)
- 6x - 14(1) = 16
- 6x - 14 = 16 ( add 14 to both sides )
- 6x = 30 ( divide both sides by - 6 )
x = - 5
solution is (- 5, 1 )
The given expression is:
(x^3 * y^4) / (3y^4)
We can notice that the term y^4 is a common term found in both the numerator and the denominator of the given expression, therefore, we can cancel this term from the both numerator and denominator (as if you divided both numerator and denominator by y^4).
Doing this, we will have the simplified form of the expression as follows:
(x^3) / (3)
The sum of the interior angles is 360
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
