Math, asked by bhavansri41056, 5 hours ago

Who answer correct and fast I will mark as brainalist...​

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Answers

Answered by tennetiraj86
2

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

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