Question

if (0.125)^2+x*16^-3y=1 and (27)^3-y*(1/3)^-2x=1,find value of x and y.plzzzz solve this....Its urgent!!!!!​

Answer 1

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9th

Maths

Linear Equations in Two Variable

Solution of Linear Equations in Two Variables

Solve the equations 2x - 3y...

MATHS

Solve the equations 2x−3y=1 and x+2y=−3 by the method of elimination

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ANSWER

Let us Solve the equations 2x−3y=1 and x+2y=−3 by the method of elimination

2x−3y=1 ......(1)

x+2y=−3 ......(2)

Multiply eqn(1) by 1 and eqn(2) by 2

2x−3y=1 ......(3)

2x+4y=−6 ......(4)

Equation (3)−(4)

2x−3y−2x−4y=1−(−6)

⇒−7y=1+6=7

∴y=

−7

7

=−1

Substituting the value of y=−1 in eqn(1) we get

2x−3×−1=1

or 2x+3=1 or 2x=1−3=−2 or x=

2

−2

=−1

Hence the co-ordinates are (−1,−1) lies in third quadrant.

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Answer 2

Answer:

first equation is unclear...but i gott the value of x=0

(1/3)^-2x=1

=>(3^2x=1

=>(3^2)^x=1

=>9^x=1

=>9^0=

=>x=0

and please tell me if it is

(0.125)^2 + x*16^(-3y)=1

or (0.125)^(2+x)    *16^(-3y)=1

Step-by-step explanation: