# Writing an algebraic equation

• Feb 28th 2007, 09:36 PM
Patience
Writing an algebraic equation
Subtract 5 from the sum of
x and 7 times 3 and the result is 20.

x-5 + 7 x 3 = 20

x = 4 is that right?
• Feb 28th 2007, 09:47 PM
Jhevon
Quote:

Originally Posted by Patience
Subtract 5 from the sum of
x and 7 times 3 and the result is 20.

x-5 + 7 x 3 = 20

x = 4 is that right?

Is that exactly how the question was phrased? if so, i'd interpret it like this:

3(x + 7) - 5 = 20
=> 3x + 21 -5 = 20
=> 3x + 16 = 20
=> 3x = 4
=> x = 4/3

I dont think the "times 3" refers to only the 7 being multiplied by 3, I'm pretty sure they'd just say 21.
• Feb 28th 2007, 09:49 PM
earboth
Quote:

Originally Posted by Patience
Subtract 5 from the sum of
x and 7 times 3 and the result is 20.

x-5 + 7 x 3 = 20
x = 4 is that right?

Hi,

your calculations are OK, but for me the following way is possible too:

First calculate the sum: (x+7)
The sum should be multiplicated by 3: (x+7)*3
Subtract 5 from this product: (x+7)*3 - 5
The result is 20:

3*(x+7) - 5 = 20 Expand the bracket:
3x + 21 - 5 = 20 Collect all constants at the RHS of the equation:

3x = 20 - 21 + 5 = 4 Divide both sides by 3:

x = 4/3

EB
• Feb 28th 2007, 10:02 PM
Patience
Yes that is the exact question.

After looking at your calculation I am kinda lost.

You say that x = 4/3

How do you solve 4/3 to = 20?
• Mar 1st 2007, 04:45 AM
Jhevon
Quote:

Originally Posted by Patience
Yes that is the exact question.

After looking at your calculation I am kinda lost.

You say that x = 4/3

How do you solve 4/3 to = 20?

By plugging it into the equation given:

3(x + 7) - 5 = 3(4/3 + 7) - 5 = 4 + 21 - 5 = 25 - 5 = 20