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Larry is 2 years younger than Mary. The difference between the squares of their ages is 28. How old is each?

2 Answers

Jan 28, 2016

Mary is $8$ $8$ ; Larry is $6$ $6$

Explanation:

Let
$XXX L$ $color(white)("XXX")L$ represent Larry's age, and
$XXX M$ $color(white)("XXX")M$ represent Mary's age.

We are told:
[equation 1] $XXX L = M - 2$ $color(white)("XXX")L=M-2$
and
[equation 2] $XXX M^{2} - L^{2} = 28$ $color(white)("XXX")M^2-L^2=28$

Substituting $M - 2$ $M-2$ from equation [1] for $L$ $L$ in equation [2]
$XXX M^{2} - {(M - 2)}^{2} = 28$ $color(white)("XXX")M^2-(M-2)^2=28$

$XXX M^{2} - (M^{2} - 4 M + 4) = 28$ $color(white)("XXX")M^2 - (M^2-4M+4)=28$

$XXX 4 M - 4 = 28$ $color(white)("XXX")4M-4=28$

$XXX 4 M = 32$ $color(white)("XXX")4M=32$

$XXX M = 8$ $color(white)("XXX")M=8$

Substituting $8$ $8$ for $M$ $M$ in equation [1]
$XXX L = 8 - 2 = 6$ $color(white)("XXX")L=8-2 = 6$

ϕ

$phi$

Jan 28, 2016

$6 and 8$ $6 and 8$

Explanation:

Let the age of $L a r r y = x$ $Larry=x$

Age of $M a r y = x + 2$ $Mary=x+2$ (Difference of their ages are 2)

Given that the difference between the squares of their ages is 28

So, ${(2 + x)}^{2} - x^{2} = 28$ $(2+x)^2-x^2=28$

Use the formula ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

$\to (4 + 4 x + x^{2}) - x^{2} = 28$ $rarr(4+4x+x^2)-x^2=28$

$\to 4 + 4 x + x^{2} - x^{2} = 28$ $rarr4+4x+x^2-x^2=28$

$\to 4 + 4 x = 28$ $rarr4+4x=28$

$\to 4 x = 28 - 4$ $rarr4x=28-4$

$4 x = 24$ $4x=24$

$x = \frac{24}{4} = 6$ $x=24/4=6$

We know now that the Age of

$L a r r y = 6$ $Larry=6$

So, Age of $M a r y = (x + 2) = 6 + 2 = 8$ $Mary=(x+2)=6+2=8$

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