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Answers
Step-by-step explanation:
Solutions :-
1)
Let the cost of one chair be Rs. X
Let the cost of one table be Rs. Y
Given that
The cost of three chairs and two tables = Rs. 700
=> 3X+2Y = 700
On multiplying with 3 both sides then
=> 9X+6Y = 2100------(1)
and
The cost of five chairs and three tables = Rs. 1100
=> 5X+3Y = 1100
On multiplying with 2 both sides then
=> 10X+6Y = 2200------(2)
on Subtracting (1) from (2)
10X+6Y = 2200
9X+6Y = 2100
(-)
____________
X+0 = 100
____________
=> X = 100
On Substituting the value of X in (1) then
9(100)+6Y = 2100
=> 900+6Y = 2100
=> 6Y = 2100-900
=> 6Y = 1200
=> Y = 1200/6
=> Y = 200
The cost of one chair = Rs. 100
The cost of two chairs = 2×100=Rs. 200
The cost of one table = Rs. 200
The cost of 3 tables = 3×200 =Rs. 600
Total cost of two chairs and three tables =
=> 200+600
=> Rs. 800
The Cost of 2 chairs and 3 tables =
Rs. 800
_____________________________
2) Given equations are :
2x+7y -5 = 0--------(1)
3x+8y = -11 ----------(2)
=>3x = -11-8y
=> x = (-11-8y)/3 ------(3)
On Substituting the value of x in (1) then
=> 2[(-11-8y)/3)]+7y -5 = 0
=> [(-22-16y)/3]+7y-5 = 0
=> (-22-16y+21y-15)/3 = 0
=> (5y-37)/3 = 0
=> 5y-37 = 0
=> 5y = 37
=> y = 37/5
On Substituting the value of y in (3)
=> x = [-11-8(37/5)]/3
=> x = (-55-296)/15
=> x = -351/15
=>x = -117/5
The solution = (-117/5,37/5)
_______________________________
3) Second equation is in complete
2x+y+z+9 is given
_____________________________
4)Given expressions are m²-3m-18 and
m²+5m+6
Now
m²-3m-18
=> m²-6m+3m -18
=> m(m-6)+3(m-6)
=> (m+3)(m-6)
and
m²+5m+6
=> m²+2m+3m +6
=> m(m+2)+3(m+2)
=> (m+2)(m+3)
now
m²-3m-18 = (m+3)(m-6)
m²+5m+6 = (m+2)(m+3)
GCD = (m+3)
GCD is the least Common factor of the prime factorization
GCD of m²-3m-18 and m²+5m+6 is (m+3)
______________________________
5) Given expression must be 3x⁴+3x³-3x-3
=> 3x³(x+1)-3(x+1)
=> (x+1)(3x³-3)
=> (x+1)(3)(x³-1)
=> 3(x+1)(x-1)(x²+x+1)
and
x³+x²-5x+3
=> x³-x²+2x²-2x-3x+3
=> x²(x-1)+2x(x-1)-3(x-1)
=> (x-1)(x²+2x-3)
=>(x-1)(x²-x+3x-3)
=> (x-1)(x(x-1)+3(x-1))
=>(x-1)(x-1)(x+3)
3x⁴+3x³-3x-3 = 3(x+1)(x-1)(x²+x+1)
x³+x²-5x+3 = (x-1)(x-1)(x+3)
GCD of the given expressions = (x-1)
GCD is the least Common factor of the prime factorization