# Solve the equations: x+3^y=3 and (x-1)^2-9^y=10 for x and y

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The equations `x + 3^y = 3` and `(x-1)^2 - 9^y = 10` have to be solved for x and y. If `3^y = z` , the system is changed to:

`x + z = 3` ...(1)

`(x - 1)^2 - z^2 = 10` ...(2)

Substitute z = 3 - x from (1) into (2)

=> `(x - 1)^2 - (3 - x)^2 = 10`

=> `x^2 - 2x + 1 - 9 - x^2 + 6x = 10`

=> 4x - 8 = 10

=> `x = 9/2`

z = -1.5

As `z = 3^y` it cannot be negative.

**The system of equations does not have a real solution.**

Given, x + 3y = 3 ------(1) and

(x-1)2 – 9y = 10 -------(2)

Taking equation (1) x + 3y = 3

Let 3y = z then, equation (1) à x + 3y= 3

=> x + z = 3

=> z = 3 – x -----(3)

Taking equation (2) ->

(x-1)2 – 9y = 10

=> (x-1)2 – (3)2y = 10

=> (x-1) 2 - ( (3)y)2 = 10

=> (x-1) 2 - z2 = 10 [ since, 3y = z ]

=> (x) 2 – 2x +1- (3 – x)2 = 10 [ substituting value of ‘z’ from equation (3) ]

=> x2 -2x +1- 9+ 6x- x2 = 10

=> 4x- 8 = 10

=> 4x = 18

=> x = 18/4 = 4.5

x = 4.5

substituting value of x in equation (1) x + 3y = 3

We get --> 4.5 + 3y = 3

=> 3y = 3 - 4.5

=> 3y = (-1.5)

Value of ‘y’ will be imaginary

**x = 4.5 and value of ‘y’ will be---->Answer**

**x = 4.5 ---->Answer**

Given, x + 3y = 3 ------(1) and

(x-1)2 – 9y = 10 -------(2)

Taking equation (1) --> x + 3y = 3

Let 3y = z then, equation (1) à x + 3y= 3

=> x + z = 3

=> z = 3 – x -----(3)

Taking equation (2) ->

(x-1)2 – 9y = 10

=> (x-1)2 – (3)2y = 10

=> (x-1) 2 - ( (3)y)2 = 10

=> (x-1) 2 - z2 = 10 [ since, 3y = z ]

=> (x) 2 – 2x +1- (3 – x)2 = 10 [ substituting value of ‘z’ from equation (3) ]

=> x2 -2x +1- 9+ 6x- x2 = 10

=> 4x- 8 = 10

=> 4x = 18

=> x = 18/4 = 4.5

**x = 4.5 ---->Answer**