Doing take aways
I am reviewing take aways. I am having trouble
How do I do $342 - 58$?
For the ones column I made the $2$ into a $12$ so I can do $12 - 8 = 4$ but I must take away one tens. So I get $3 - 5$ in tens column but I cant do $3 - 5$. What do I do now do I borrow something else?
$$ \begin{array}{rrrr} &3&4&2\\ -& &5&8\\\hline \end{array} $$
We can't take 8 from 2, so we "borrow" one of the 4 tens from 342, and make it into 10 units instead, and add it to the 2 units we already have. That leaves 3 tens in the tens' column, and 12 units in the units' column:
$$ \require{cancel} \begin{array}{rrrr} \quad &3&\color{darkred}{\cancelto{3}4}&\color{darkred}{{}^1}2\\ -& &5&8\\\hline \end{array} $$
What we did here is like making change: We had 4 dimes and 2 pennies, but that was not convenient because we wanted to take away 8 pennies. So we changed one of the dimes into pennies, and then we had 3 dimes and 12 pennies, and it was easy to take away 8 pennies. This "borrowing" trick is just like making change.
Then we take 8 units from 12 units, leaving 4 units: $$ \require{cancel} \begin{array}{rrrr} \quad &3&{\cancelto{3}4}&{{}^1}2\\ -& &5&8\\\hline &&&\color{darkred}{4}\end{array} $$ So far, this is exactly what you did.
Now we want to subtract 5 tens from 3 tens, but, as you observed, we can't. So we do almost the same as before, but this time we "borrow" one of the hundreds from the 3 hundreds in the hundreds' column, and turn it into 10 tens, adding it to the 3 tens that are already in the tens' column:
$$ \require{cancel} \begin{array}{rrrr} \quad &\color{darkred}{\cancelto23}&\cancelto{\color{darkred}{1}3}4&{}^12\\ -& &5&8\\\hline &&&4 \end{array} $$
Again, this is like making change: we had 3 dollars and 3 dimes, and we changed one dollar into ten dimes, leaving 2 dollars and 13 dimes. That is allowed in arithmetic for the same reason it is allowed in money dealings: it doesn't change the actual amount!
Anyway subtracting 5 tens from 13 tens we have 8 tens:
$$ \require{cancel} \begin{array}{rrrr} \quad &\cancelto23&\cancelto{13}4&{}^12\\ -& &5&8\\\hline &&\color{darkred}{8}&4 \end{array} $$
and then 2 hundreds minus no hundreds is 2 hundreds:
$$ \require{cancel} \begin{array}{rrrr} \quad &\cancelto23&\cancelto{13}4&{}^12\\ -& &5&8\\\hline &\color{darkred}{2}&8&4 \end{array} $$
And the answer is 284.
I've personally never been good at doing this with the method they teach in the third grade (in fact, I can't even remember it, and I'm a math student in university!). I prefer to do it like this, taking away a little bit at a time.
You need to take away $58$ from $342$.
Well, first let's take away $42$, because that's easy. $342-42$ is just $300$.
Okay, now how much more do we have to take away? We already took away $42$, so how much more to get to $58$? Well, if I add $8$ to $42$ I get $50$, so I need $8$ more again to get to $58$, so I need $8 + 8$ overall. That's $16$. So I need to take away $16$ more.
So how do I take away $16$ from $300$? $16$ is a $10$ and a $6$, so let's do those one a time. Taking away $10$ is easy. $300-10$ is $290$. Now taking away that last $6$ isn't so hard, you can do it with your fingers if you like. $290-6$ is $284$.
So the answer is $342-58=284$.
So the trick is basically:
- Take away a little bit whch is easy to take away.
- Figure out how much there is left to take away.
- Repeat until you're done.
Of course, maybe your school wants you to learn it in their specific way, but if you find the method I used easier, maybe that'll at least help.
You lower the hundreds spot by one, and increase the tens spot by 10. Example:
300 + 40 + 2 = 200 + 140 + 2
Now that you do this, you get 13-5 and your answer is an 8 in the tens place, and with a 2 in the hundreds place.
As a whole, it looks like this
342 - 58
300 + 40 + 2 - 50 - 8
200 + 130 + 12 -50 - 8 (1 hundred turns into 10 tens, and 1 ten turns into 10 ones)
200 + 130 -50 + 12 - 8 (just putting them in order)
200 + 80 + 4
284 is the answer
While some answers here have been doing it in different, and maybe more efficient methods, I'd like to explain as it seems like your teacher (probably) wants you to do it :)
When doing $342 - 58$, you start with $2 - 8$. This does not work, so you have to borrow a ten. $12 - 8 = 4$. This is the last digit, your ones. Now, since we borrowed a ten, we have to do $3 - 5$. We have to borrow from the first number, and we get $13 -5 = 8$. This is the second digit; your tens. Now, since we are left with $2$ after we borrowed from the three, we have $284$, or "two hundreds, eight tens and four ones."
Best of luck with catching up, I know how hard it is to come from behind in math :)
If you are having trouble setting it up in a table, MJD's answer will help!