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Find 4 consectutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers.

Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers.
I'm never able to understand word problems...

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Brad M. | Professional Tutor: Ultra Streamlined Math - Physics - AccountingProfessional Tutor: Ultra Streamlined M...
4.9 4.9 (216 lesson ratings) (216)
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Hey Veronica -- here's an intuitive approach ... since the difference is 72,

we need at least 10X8 ...

subtracting the 6x4 leaves 80-24=56 too small ... try the next set up

==> 12x10 minus 8x6= 120-48=72 ... 12,10,8,6  Best wishes, ma'am :)

Kristina K. | Stuy Grad, Dean's List, 8+ Years Experience, K-8, SHSAT, SATStuy Grad, Dean's List, 8+ Years Experie...
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First number: x

Second number: x+2

Third number: x+4

Fourth number: x+6

 

Product of two smaller numbers: (x)(x + 2) = x^2 + 2x

Product of two larger numbers: (x+4)(x+6) = x^2 + 6x +4x + 24 = x^2 +10x +24

 

product of the two smaller numbers is 72 less than the product of the two larger numbers:

x^2 + 2x = x^2 + 10x + 24 -72

So group your variables and you get: 

-8x = -48

x= 6

since you know x = 6, you know the even numbers that will come after it.

x = 16.5

 

 

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Ralph L. | Algebra I, II, Visual Basic, Beginning C++ tutorAlgebra I, II, Visual Basic, Beginning C...
4.0 4.0 (1 lesson ratings) (1)
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x * (x +2) = (x + 4) * (x + 6) - 72

x2 + 2x = x2 + 10x + 24 - 72

subtract x2 from both sides:

2x = 10x - 48

2x - 10x = -48

-8x = -48

x = 6

so, consecutive even integers are:

6, 8, 10, 12