hi how would I work out the following?
(3(2^2 -2)^2 -10)^2
which brackets do I calculate first please?
HI again .
If you need any more help , its better to start a new thread . Well , i am going to answer this . If you have any more q's , start a new onw
$\displaystyle \frac{2x-1}{x+2}=3$
$\displaystyle 2x-1=3(x+2)$
$\displaystyle 2x-1=3x+6$
$\displaystyle x=-7$
$\displaystyle
x^2+3x-9=(x-4)(x+21)
$
$\displaystyle x^2+3x-9=x^2+17x-84$
$\displaystyle 3x-9=17x-84$
$\displaystyle 14x=75$
$\displaystyle x=75/14$
So 5 is not a solution .
You seem to have the wrong sign for both of them
$\displaystyle \frac{2(7)-1}{7+2} = \frac{13}{9} \neq 3$
by inspection $\displaystyle x = -7$
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For the 2nd one $\displaystyle 31 \neq 26$ so 5 is not a solution
$\displaystyle x^2+3x-9 = x^2+17x-84$
$\displaystyle -14x = - 75$