How do you evaluate $x^2/y + x^2/z$ for $x=6, y=-4,$ and $z=-2$ ?

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Answer 1 · 2016-07-19T18:04:25

The answer will be $-27$ .

Simply plug in the values and simplify:

$(6^2)/-4 + (6^2)/-2$

$=36/-4 + 36/-2$

$=-9 - 18$

$=-27$

Answer 2 · 2016-07-19T18:08:09

$-27$ .

$x^2/y+x^2/z=x^2(1/y+1/z)=(x^2(y+z))/(yz)=(6^2(-4-2))/((-4)(-2))=(36(-6))/8=-27$ .

How do you evaluate x^2/y + x^2/zx^2/y + x^2/z for x=6, y=-4,x=6, y=-4, and z=-2z=-2?