I'm taking math 051 at a community college and need help with an integer problem...please help?

Question

Here is the problem: The sum of two integers is 10. Three times the larger integer is three less than eight times the smaller integer. find the integers

could you show work please so I can practice!!!!

Answer 1

"The sum of two integers is 10" is: x + y = 10
"Three times the larger integer is three less than eight times the smaller integer" is 3x = 8y - 3 (where x is the larger integer and y is the smaller integer).

From the first statement, we can put x in terms of y: x = 10 - y

Plug "10 - y" for x in the second statement:

3(10 - y) = 8y - 3
30 - 3y = 8y - 3
33 = 11y
3 = y

Plugging the value of y into the first statement:

x + 3 = 10
x = 7

Your integers are 7 and 3.

Answer 2

Let x and y be the two integers, and let y be the larger integer.

Then,

x + y = 10 ("The sum of two integers is 10.")

3y = 8x - 3 ("Three times the larger integer IS three less than 8 times the smaller integer.")

We have two equations, and two unknowns:

x + y = 10
3y = 8x - 3

Let's rearrange 3y = 8x - 3 so that it's in the same form as the other equation

x + y = 10
8x - 3y = 3

To solve this, we use substitution.

x + y = 10 means y = 10 - x.
Plugging in y = 10 - x into the second equation,

8x - 3(10 - x) = 3

Expand by distributing the -3 over the brackets,

8x - 30 + 3x = 3

Group like terms; move the -30 to the right hand side.

11x = 33

Divide both sides by 11, which gives you x = 3.

Now that you know x = 3, you can obtain y.
y = 10 - x, so y = 10 - 3 = 7

Your two integers are x = 3, y = 7.

Answer 3

You need to set up two simultaneous equations. Assign variables to both

x = larger integer
y = smaller integer

Your first equation should look like this
X + Y = 10

Your second one will look like this
3X = 8Y - 3

Three times the larger integer = 3X
3 less than 8 times the smaller integer = 8Y - 3

Now rewrite the equation so that both variables are on one side

3X - 8Y = -3


You now have two equations to solve:
1) X + Y = 10
2) 3X - 8Y = -3

Multiply equation 1 by negative 3. It now looks like
-3X + -3Y = -30

Add this equation to equation 2. This will cancel out the X's leaving you with only one variable

-11Y = -33

Divide both sides by -11

Y = 3

Plug this back into equation number one and solve for X

X = 7

Answer 4

Write down all the facts
x + y = 10
8y - 3x = 3

isolate a variable
x = 10 - y

substitute the variable u isolated
8y - 3(10 - y) = 3
8y - 30 + 3y = 3
11y = 33
y = 3

therefore x = 7

x = 7; y = 3

Answer 5

2 integers x and y

x+y=10 =>x=10-y
Assume x largest: 3*x+3=8*y

=>3*(10-y)+3=8*y=>30-3*y+3=8*y
=>11*y=33
=>y=3 and x=10-3=7

Answer 6

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