Questions in introductory algebra

1. Solve the equation x + 5 = 19

2. Find two numbers that add to 6 and multiply to 8.

3. Sam paid his bill of $9.85 with only quarters, dimes
and nickels.
He used twice as many quarters as nickels, and more nickels
than dimes.
How many quarters, dimes, nickels did he use?


4. Sam paid his bill of $9.85 with only quarters, dimes and
nickels.
He used twice as many quarters as nickels, and one less
nickel than dimes.
How many quarters, dimes, nickels did he use?

5. Sam paid his bill of $9.85 with only quarters, dimes and
nickels.
He used twice as many quarters as nickels, and one less
quarter than dimes.
How many quarters, dimes, nickels did he use?


Question # 1
Solve the equation x + 5 = 19

The equation x + 5 = 19 means
there is a number that we name x,
and when we add 5 to that number, we get 19.

The solution is to subtract 5 from 19.
( x + 5) - 5 = 19 - 5
x + 5 - 5 = 14
x = 14

Verify that the answer is correct.
14 + 5 = 19
19 = 19

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Question #2
Find two numbers that add to 6 and multiply to 8.

It's easy to think of the answer to this, without using
algebra.

But since this is an algebra question, let's use algebra to
answer the question.

Let x and y be the names of the two numbers.
Then
x + y = 6
x * y = 8

solve the first equation for y.

y = 6 - x

Plug into the second equation.
x * (6 -x) = 8
6 * x - x**2 = 8

Move all the terms to one side of the equation,
changing signs of terms that cross the equal sign.
x**2 - 6 * x + 8 = 0

This is a quadratic equation that can be solved, either by
factoring, or by using the quadratic formula.

Either method gives solutions x = 2 or x = 4.
So take one of 2 or 4 to be x, and the other to be y.
The two numbers are 2 and 4.

Verify that the answer is correct.
2 + 4 = 6
2 * 4 = 8

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Question # 3

Sam paid his bill of $9.85 with only quarters, dimes
and nickels.
He used twice as many quarters as nickels, and more nickels
than dimes.

How many quarters did he use?

How many dimes did he use?

How many nickels did he use?

Let q be the name of the number of quarters.
Let d be the name of the number of dimes.
Let n be the name of the number of nickels.

25 q + 10 d + 5 n = 985
q = 2 n
n > d

25 ( 2 n) + 10 d + 5 n = 985
50 n + 10 d + 5 n = 985
55 n + 10 d = 985
11 n + 2 d = 197
197 has a remainder of 10 when divided by 11
11 * 17 + 10 = 197
11 * 17 + 2 * 5 = 197
11 n + 2 d = 197
n > d

n could be 17 and d could be 2 since 17 > 2.

Could n be 16?

if n is 16, 11 * 16 + 2 d = 197
2 d = 197 - 11 * 16 = 197 - 176 = 21.
n cannot be 16 because 21 is odd.
could n be 15?
if n is 15, 11 * 15 + 2 d = 197

2 d = 197 - 11 * 15 = 197 - 165 = 32
d = 16.
n cannot be 15 because then d would be 16, which is > 15.

n = 17, 2 d = 197 - 11*17 = 197 - 187 = 10
d = 5
q = 2*n = 2 * 17 = 34

Check answer.
34 * 25 + 5 * 10 + 17 * 5
= 850 + 50 + 85 = 985

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Question # 4. Sam paid his bill of $9.85 with only quarters, dimes and
nickels.
He used twice as many quarters as nickels, and one less
nickel than dimes.

How many quarters did he use?

How many dimes did he use?

How many nickels did he use?

25 q + 10 d + 5 n = 985
q = 2 * n
n = d - 1
d = n + 1

25 * (2 * n) + 10 * (n + 1) + 5 n = 985
50 * n + 10 * n + 10 + 5 * n = 985
(50 + 10 + 5) * n + 10 = 985
65 * n = 975
13 * n = 195
n = 15

d = n + 1 = 16
q = 2 * n = 30
Check answer.
30 * 25 + 16 * 10 + 15 * 5 = 750 + 160 + 75 = 985

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Question # 5.
Sam paid his bill of $9.85 with only quarters, dimes and
nickels.
He used twice as many quarters as nickels, and one less
quarter than
dimes.

How many quarters did he use?

How many dimes did he use?

How many nickels did he use?

25 q + 10 d + 5 n = 985
q = 2 * n
d = q + 1 = 2 * n + 1
25 * (2 * n) + 10 * ( 2 * n + 1) + 5 * n = 985
50 * n + 20 * n + 10 + 5 * n = 985
(50 + 20 + 5) * n = 985 - 10 = 975
75 * n = 975
15 * n = 195
3 * n = 39
n = 13
q = 2 * n = 26
d = q + 1 = 27

Test correctness of answer.

26 * 25 + 27 * 10 + 13 * 5 = 650 + 270 + 65 = 985

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