Basic Algebra Word Problem...PLEASE HELP!!

Question

A cashier has 50 $20 bills and $5 bills in her cash drawer. The value of the bills is $535. How many $20 bills are there?

Answer 1

If you get a problem like this on a timed test, take the quick way out. Let T stand for how many twenties and F stand for how many fives. The amount of money is how many times how much each bill is worth.

T + F = 50

20T + 5F = 535

Multiply each term in the top equation by -5 to get rid of the Fs. Subtract it from the bottom one.

20T + 5F = 535

-5T - 5F = -250

________________

15T = 285

T = 19

Stop right here since you're not asked for how many fives.

Answer 2

Let's let x represent the NUMBER of $20 bills and y represent the NUMBER of $5 dollar bills. Since there are a total of 50 bills , we can write:

x + y = 50

Then since each $20 bill is worth twenty dollars or $20 and each $5 bill is worth five dollars or $5, we can write:

20x + 5y = $535

To solve these two equations with two unknowns, use substitution.

From the first equation we can write that y= 50 - x

Substitute this expression for y into the second equation:

20x + 5(50 - x ) = 535

Solve for q using inverses operations, combining like terms, and the distributive property:

20x + 250 - 5x = 535

15x + 250 = 535

15x + 250 - 250 = 535 - 250

15x = 285

15x /15 = 285/15

x=19

So the number of $20 bills is 19 . The number of $5 bills will be 50 -19 or 31 $5 bills.

To check, 19 twenty dollar bills would be worth $380 (OR 20 * 19)

31 five dollar bills would be worth $155 (OR 5 * 31)

$380 + 155 = $535