Find two consecutive even integers such that three times the lesser integer, added to the larger integer, is 58.
Even integer = 2x Next even integer = 2x + 2 Now, with the problem equation: 3*(2x) + 2x+2=58 6x+2x=56 8x=56 x=7 So, the integers are 2x and 2x+2, 14 and 16.
We can let even number #1 = 2n and we can let even number #2 = 2n + 2 so we have...3 times the lesser 3(2n) added to the larger (2n + 2) = 58 so 3(2n) + 2n + 2 = 58 solve for 'n'
The two consecutive two numbers are : \[a=2n~~\text{ et }~~ b=a+2=2n+2\] And we have : \[3a+b=58\] So : \[6n+2n+2=58\] So : \[8n=56\] So : \[n=7\] So : The two consecutive two numbers are : \[a=14~~~ b=16\]
lets see what happens if we let one simply equal \(n\) and the other \(n+2\)
then solve \[3n+(n+2)=58\] \[4n+2=58\] \[4n=56\] \[n=14, n+2=16\] don't let that "even" business throw you off
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