Check my Work: Factoring for 2y^+9y-18 is (2y-3) (y+6) true or false?

1 answer:

7 0

It is true because when you separate 2y^2 +9y -18, you are supposed to get the factors (2y-3)(y+6), when you multiply this it gives you the first equation, we know this because when we multiply the first term it gives us exactly 2y^2 and if we multiply the last ones it gives you -18, now here is the tricky part, when we multiply the middle part we get 12y and -3y, when we add these up we get 9y, now is that the same middle as in the equation? Yes… then our factors are correct making this true

Hope this helps

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$Step \; 1: \; Assing \; variable \; for \; the \; unknown \; that \; we \; need \; to \; find.\\Let \; x \; be \; additional \; money \; invested$

$Step \; 2: \; Use \; the \; appropriate \; formula \; of \; simple \; interest \; set \; up \; an \; equation\\Interest \; earned \; each \; year = \; Amount \; invested \times annual \; rate \; of \; interest\\\\Amount \; invested \; in \; 4 \% \; rate \; of \; interest=\$4500\\Amount \; invested \; in \; 5\frac{1}{2}\% \; rate \; of \; interest= \$x\\Amount \; invested \; totally \; in \; 4\frac{1}{2}\% \; rate \; of \; interest = \$(4500+x)$

$Interest \; earned \; in \; 4\% \; interest \; rate\\ = \; 4500(4\%)=180\\Interest \; earned \; in \; 5\frac{1}{2}\% \; interest \; rate\\ = x(5\frac{1}{2}\%)=0.055x\\Interest \; earned \; totally \; in \; 4\frac{1}{2}\% \; interest \; rate = (4500+x)(4\frac{1}{2}\%)\\=0.045(4500+x)$

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$\underline{Conclusion:}\\Ben \; should \; invest \; additional \; money \; of \; \$2250$

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