Three cousins have ages that are consecutive integers. The product of the two older counsins ages is twelve less than six times the sum of the younger two cousins ages. Write an equation and solve to find the three cousins ages algebraically.

n, n+1, n+2

(n+1)(n+2)=6(n+n+1)-12

n^2+3n+2=12n+6-12

n^2+3n+2=12n-6

n^2-9n+8=0

n^2-n-8n+8=0

n(n-1)-8(n-1)=0

(n-8)(n-1)=0

So there are actually two solutions that satisfy the conditions for the cousin's ages...

1,2, and 3 years old and 8,9,and 10 years old.

check...

2*3=6(1+2)-12, 6=18-12, 6=6

9*10=6(8+9)-12, 90=102-12, 12=12

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ap8997154 ap8997154 · Answer 2 · 2022-04-12T14:55:14+02:00

An equation is formed of two equal expressions. The age of the cousin can be either 1, 2, and 3 or 8, 9, and 10.

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The age of the three cousin brothers is consecutive integers. Therefore, the let the ages of the three cousins be x, (x+1), and (x+2).

Now, according to the statement, the product of the two older cousins' ages is twelve less than six times the sum of the younger two cousins' ages. Therefore, the equation can be written as,

[tex](x+1)(x+2)=6[x+(x+1)]-12\\\\x^2+3x+2=6(2x+1)-12\\\\x^2+3x+2=12x+6-12\\\\x^2+3x-12x+2+12-6=0\\\\x^2-9x+8=0\\\\x^2-8x-x+8=0\\\\(x-8)(x-1)=0[/tex]

Now, since the equation has two solutions, the solution can be either 1, 2, and 3 or 8, 9, and 10.

Let substitute the value of x in the above equation, to check if the solution is a valid answer is not.

When x = 1,

[tex](x+1)(x+2)=6[x+(x+1)]-12\\\\(1+1)(1+2)=6[1+(1+1)]-12\\\\2 \times 3 = 6[1+2]-12\\\\6=18-12\\\\6=6[/tex]

When x=8,

[tex](x+1)(x+2)=6[x+(x+1)]-12\\\\(8+1)(8+2)=6[8+(8+1)]-12\\\\9 \times 10 = 6[8+9]-12\\\\90=102-12\\\\90=90[/tex]

Thus, the age of the cousin can be either 1, 2, and 3 or 8, 9, and 10.

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