Not so radical expression

Algebra Level 1

y 2 + 10 y + 24 y 2 + 14 y + 48 ÷ y 2 − 14 y + 48 y 2 + 2 y − 48 = ? \large \frac {y^2 + 10y + 24}{y^2 + 14y + 48} \div \frac {y^2-14y +48}{y^2+2y-48} = \ ?

y 2 + 4 y + 16 y \frac{y^2+4y+16}{y} y + 4 y − 8 \frac{y+4}{y-8} y 3 y^3 y 6 \frac{y}{6}

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Diana Wang by Diana Wang , 20, China

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1 solution

Nihar Mahajan
Feb 4, 2015

The given expression can be written as-

( y + 4 ) ( y + 6 ) ( y + 8 ) ( y + 6 ) ÷ ( y − 8 ) ( y − 6 ) ( y − 6 ) ( y + 8 ) \displaystyle\frac{(y+4)(y+6)}{(y+8)(y+6)}\div\frac{(y-8)(y-6)}{(y-6)(y+8)}

= ( y + 4 ) ( y + 6 ) ( y + 8 ) ( y + 6 ) × ( y − 6 ) ( y + 8 ) ( y − 8 ) ( y − 6 ) \displaystyle\ =\frac{(y+4)(y+6)}{(y+8)(y+6)}\times\frac{(y-6)(y+8)}{(y-8)(y-6)}

Terms like ( y + 6 ) , ( y + 8 ) , ( y − 6 ) (y+6) , (y+8) , (y-6) get cancelled

= y + 4 y − 8 =\boxed{\displaystyle\frac{y+4}{y-8}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Correct. For completeness, you should show for what values of y y can't this expression be simplified to.

Just to be absolutely precise, one must note that the denominator can't be zero. ∴ y ≠ ± 6 . . . . a n d y ≠ ± 8 \therefore\ y \ne \pm 6 \ .... and \ y \ne \pm 8 Otherwise it's a perfect solution

Curtis Clement - 6 years, 1 month ago

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