Thread: Exponential simultaneous equations

Help with these would be nice =)

Solve the simultaneous equations:
2^x - 3^y = 7
2^x-1 + 3^y+1 = 35

And another question:
If 2,1 is a solution to the equations: x^2 + xy + ay = b, and 2ax + 3y = b, find the other solution.

I did:
4 + 2 + a = b --(1)
4a + 3 = b -- (2)
4a + 3 = 6 + a
a = 0.6
b = 6.6
Then I substituted it into the original equations to get:
x^2 + xy + 0.6y = 6.6 -- (3)
1.2x + 3y = 6.6 --(4)

But I can't get anywhere with it...

Originally Posted by erika

Solve the simultaneous equations:
2^x - 3^y = 7
2^x-1 + 3^y+1 = 35

I presume the second equation is actually:
2^{x - 1} + 3^{y + 1} = 35

Please use parenthesis!!

We know that 3^y = 2^x - 7

So...

2^{x - 1} + 3^{y + 1} = 35

(1/2)*2^x + 3*3^y = 35

(1/2)*2^x + 3*[2^x - 7] = 35

(1/2)*2^x + 3*2^x - 21 = 35


(7/2)*2^x = 56

2^x = (2/7)*56 = 16

2^x = 2^4

Thus x = 4 and 3^y = 2^4 - 7 = 16 - 7 = 9

So y = 2

-Dan

Originally Posted by erika

Help with these would be nice =)

Solve the simultaneous equations:
2^x - 3^y = 7
2^x-1 + 3^y+1 = 35

And another question:
If 2,1 is a solution to the equations: x^2 + xy + ay = b, and 2ax + 3y = b, find the other solution.

I did:
4 + 2 + a = b --(1)
4a + 3 = b -- (2)
4a + 3 = 6 + a

You are good to this point.

4a + 3 = 6 + a

3a = 3

Thus a = 1 and b = 6 + a = 7

So:
x^2 + xy + y = 7
2x + 3y = 7

Solve the bottom equation for y:
y = (-2/3)x + (7/3)

Thus
x^2 + x[(-2/3)x + (7/3)] + [(-2/3)x + (7/3)] = 7

x^2 + (-2/3)x^2 + (7/3)x + (-2/3)x + (7/3) = 7 <-- Multiply both sides by 3

3x^2 - 2x^2 + 7x - 2x + 7 = 21

x^2 + 5x - 14 = 0

(x - 2)(x + 7) = 0

Thus x = 2 or x = -7

For x = 2 ==> y = (-2/3)(2) + (7/3) = -4/3 + 7/3 = 1 (as expected)

For x = -7 ==> y = (-2/3)(-7) + (7/3) = 14/3 + 7/3 = 21/3 = 7

So your other solution is (-7, 7).

-Dan

For the first question:
if x = 4 and y= 9,
2^x - 3^y
= 2^4 - 3^9
= -19667 (if i didn't see it wrongly?)

Hello, Erika!

Dan did a great job . . . just slipped on the very last step.

Solve the simultaneous equations:
. . [1] .2^x - 3^y .= .7
. . [2] .2^{x-1} + 3^{y+1} .= .35

Equation [2] is: .½·2^x + 3·3^y .= .35

Multiply by 2: .2^x + 6·3^y .= 70
.Subtract [1]: .2^x .- . 3^y . = .7

and we have: .7·3^y .= .63 . . → . . 3^y = 9 . . → . . y = 2

Substitute into [1]: .2^x - 3² .= .7 . . → . . 2^x = 16 . . → . . x = 4

Thanks Dan and Soroban.
I think there may be a problem with the 2nd question too though:

The question states: 2x + 3y = 7,
but using the figures from the answer:
2(-7) + 3(7/3) = -7

Sorry about this, I really appreciate the help =)

Hello, Erika!

If (2,1) is a solution to the equations: x² + xy + ay .= .b,
and: 2ax + 3y .= .b, find the other solution.

I did:
4 + 2 + a .= .b . . (1)
. . 4a + 3 .= .b . . (2)

. . 4a + 3 .= .6 + a

Then: .a = 0.6 . . . . no

You have: .3a .= .3 . . → . . a = 1 . . → . . b = 7

The equations become: .x² + xy + y .= .7 . [1]
. . . - . . - . . . . . . . . . . . 2x .+ .3y . = .7 . [2]

Solve [2] for y: .y .= .(7 - 2x)/3

Substitute into [1]: .x² + x(7 - 2x)/3 + (7 - 2x)/3 .= .7

Multiply by 3: .3x² + 7x - 2x² + 7 - 2x .= .21

We have the quadratic: .x² + 5x - 14 .= .0

. . which factors: .(x - 2)(x + 7) .= .0

. . and has roots: .x .= .2, -7

. . and we have: .y .= .1, 7


We were given the solution (2,1).

. . The other solution is (-7,7).