Find three consecutive even integers that sum up to -72.

Mathematics

1 answer:

ozzi3 years ago

5 0

Answer:

Step-by-step explanation:

$n,\ n+2,\ n+4-\text{three consecutive even integers}\\\\\text{The equation:}\\\\n+(n+2)+(n+4)=-72\\\\n+n+2+n+4=-72\qquad\text{combine like terms}\\\\3n+6=-72\qquad\text{subtract 6 from both sides}\\\\3n+6-6=-72-6\\\\3n=-78\qquad\text{divide both sides by 3}\\\\\dfrac{3n}{3}=-\dfrac{78}{3}\\\\n=-26\\\\n+2=-26+2=-24\\\\n+4=-26+4=-22$

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3 years ago

I need help on number 16 these instructions say to solve these equations by cross multiplying

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Answer:

15. x = 4

16. a = 19

Step-by-step explanation:

"Cross multiply" essentially means "multiply both sides of the equation by the product of the denominators."

$\dfrac{x}{6}=\dfrac{2}{3}\\\\3(x)=6(2)\qquad\text{"cross multiply"}\\\\\boxed{x=4}\qquad\text{divide by 3}$

This is an example of a problem in which cross-multiplying introduces the need for an extra step: divide by 3. Had we simply multiplied by the denominator of the x-term, we would have been finished in one step:

x = 6(2/3) = 4

That is, rather than multiplying by the <em>product of the denominators</em>, sometimes the problem can be solved more easily if you multiply by their <em>least common multiple</em>.

$\dfrac{a+1}{a+5}=\dfrac{a-4}{a-1}\\\\(a-1)(a+1)=(a+5)(a-4)\qquad\text{"cross multiply"}\\\\a^2-1=a^2+a-20\qquad\text{simplify}\\\\\boxed{19=a}\qquad\text{add $20-a^2$}$