# Thread: AS question on powers, sim eq, and inequalities

1. ## AS question on powers, sim eq, and inequalities

These are some questions from a AS Core 1 Practise Paper and I don't have a clue what to do.I had a go at B a few times and got a few different answers based around root7.

A) Given that 3^x = 9^(y-1) , show that x=2y-2.

B) Solve the simulatious eqs, x=2y-2
x^2=(y^2)+7

C) Find a set of values for x for which 6x-7 < x+3 and (2x^2)+7x -4 >0

I had a go at C and solved each equation, x<2 and -4<x<1/2 but wasn't sure what to do when it says find a set of values for both.

Sorry I'm not sure how to write powers properly on computors
Thanks

2. Originally Posted by markcantdomaths
A) Given that 3^x = 9^(y-1) , show that x=2y-2.
Note that $\displaystyle 9^{y - 1} = \left( {3^2 } \right)^{y - 1} = 3^{2(y - 1)} .$

Originally Posted by markcantdomaths
B) Solve the simulatious eqs, x=2y-2
x^2=(y^2)+7
Substitute the first equation into the second. You'll get a quadratic equation that you should be able to tackle.

Originally Posted by markcantdomaths
C) Find a set of values for x for which 6x-7 < x+3 and (2x^2)+7x -4 >0

I had a go at C and solved each equation, x<2 and -4<x<1/2 but wasn't sure what to do when it says find a set of values for both.
Did you do a sketch? That'd be helpful.

3. Oh that was actually quite simple, just the x and y powers scared me! Thanks.

4. Originally Posted by markcantdomaths
These are some questions from a AS Core 1 Practise Paper and I don't have a clue what to do.I had a go at B a few times and got a few different answers based around root7.

A) Given that 3^x = 9^(y-1) , show that x=2y-2.

B) Solve the simulatious eqs, x=2y-2
x^2=(y^2)+7

...
Hello,

to A):

$\displaystyle 3^x = 9^{y-1}~\iff~3^x = \left( \left(3 \right)^2 \right)^{y-1}~\iff~3^x = 3^{2x-2}$

Since the bases are equal the exponents must be equal too. Thus

$\displaystyle x = 2y-2$

to B) Plug in the term for x into the 2nd equation:

$\displaystyle (2y-2)^2 = y^2 + 7~\implies~ 4y^2-8y+4=y^2+7~\implies~ 3y^2-8y-3=0$

Solve this quadratic equation for y. I've got:

$\displaystyle y = 3~\vee~y = -\frac13$ Plug in these values into the first equation to calculate x:
$\displaystyle x = 4~\vee~x=-\frac83$