find two consecutive odd numbers such that the sum of three sevenths of the first numberand one third of the second numberis equal to 38

So you want \(\frac{3}{7}x + \frac{1}{3}y = 38\), where x and y are consecutive odd numbers. The fact that they are consecutive odd numbers means that the relationship between the two can be described as \(x = y + 2\) (or vice versa). That means you have: $$ x = y + 2 $$ $$ \frac{3}{7}x + \frac{1}{3}y = 38 $$ Can you figure out how to solve this using systems of equations?

im confused

hmmm, well in this question you have two unknowns, the first number and the second number

(let's call them x and y). according to the second part of your prompt, we know 3x/7 + y/3 = 38 - that's what shadowfiend wrote too

now, given that the numbers are consecutive, odd numbers, you have another equation you can write too

since all consecutive odd numbers are two numbers apart (since every other number is even), you know the second number (y) is y = x + 2

if you substitute that in for y in the first equation, you should have the answer

or rather, you'll be able to solve for x

and then consequently you'll be able to get y (by substituting the actual value of x into either equation)

thank you :)

yep no problem, the good thing is once you think you've solved these, they're easy to check, since you just put the numbers back into your equations and make sure the statements are true

Thanks for covering, sandra, great explanation :)