Question 6018c

Dec 12, 2017

$356 - 293 = 63$ , or "6 tens and 3 ones".

$49 \times 10 + 82 \times 1 - \left(12 \times 10 + 96 \times 1\right) - \left(6 \times 10 + 3 \times 1\right) = 293$

Explanation:

Using the place descriptions, the equation is:

$49 \times 10 + 82 \times 1 - \left(12 \times 10 + 96 \times 1\right) = 293$

$490 + 82 - \left(120 + 96\right) = 293$
$572 - 216 = 293$
$356 = 293$ SO the "missing factor" on the other side must be

$356 - 293 = 63$ , or "6 tens and 3 ones".

$49 \times 10 + 82 \times 1 - \left(12 \times 10 + 96 \times 1\right) - \left(6 \times 10 + 3 \times 1\right) = 293$

Dec 12, 2017

The value of $x$ is $63$
You can express this amount as "6 tens 3 ones."

Explanation:

I think the question has a typo where the $x$'s didn't show up.

I think the question should be:
If 49 tens 82 ones - 12 tens 96 ones = $x$ + 293, then find the value of $x$

This question implies that "293" is too small, and it wants to know how much you'd need to add to make the answer correct.

Here is a good way to solve this problem.

1) For your own convenience, turn the bills into a sum of money.

49 tens  = $490 82 ones =$   82
......................................
Total . . . . . $572 12 tens =$120
96 ones = $96 ..................................... Total. . . . . .$216

Now you can write the question like this:
[49 tens 82 ones] $-$ [12 tens 96 ones] $=$ $x$ + 293
[ . . . . . 572 . . . . . ] $-$ [ . . . . . 216 . . . . . ] $=$ $x$ + 293

$572 - 216 = x + 293$
Solve for $x$

1) Combine like terms by doing the subtraction
$356 = x + 293$

2) Subtract 293 from both sides to isolate $x$
$63 = x \leftarrow$ answer

The value of $x$ is $63$
$572 - 216$ . . $=$ . . . 63 + 293?
. . . . $356$ . . . . .$=$ . . . . .$356$
Check!#