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 Post subject: Book 2 - Numeber 18 pg 31Posted: Fri Dec 02, 2011 7:24 pm

Joined: Fri Mar 18, 2011 12:27 pm
Posts: 17
Hello

I understand the book's explaination right up to the last paragraph on page 32.
20x=3y and 11x + 7 = 2y but how do you solve those equations to get x = 3 and y = 20?

Thank you for your help!

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 Post subject: Re: Book 2 - Numeber 18 pg 31Posted: Mon Dec 05, 2011 1:18 pm

Joined: Thu Dec 01, 2011 6:36 am
Posts: 6
While I am having trouble with the first part of the question, I may have solved this through numbers...

Anyways, for solving the equations, this is what I would do:
20x=3y, hence y = 20x/3, substitute that in the other equation you get:
11x+7=2y or 11x+7=2*20x/3 or 11x+7=40x/3
multiply both sides by 3:
33x+21=40x

subtract 33x from both sides:
21=7x or x = 21/7 or x = 3

then use x = 3 in the previous equation y=20x/3:
y=20*3/3 or y = 20...

Hope that helped, you may try equation subtraction as well...

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 Post subject: Re: Book 2 - Numeber 18 pg 31Posted: Mon Dec 05, 2011 8:30 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Great job solving!

You referenced combining equations to solve as well:

If you'd like to try adding / subtracting equations, this is an approach:
Equation 1: 20x=3y
Equation 2: 11x + 7 = 2y

Take Equation 1 and multiply by 2 and then take Equation 2 and multiply by 3. (We want the same number of y's in both of them)

Eq 1 becomes: 40x = 6y
Eq 2 becomes 33x + 21 = 6y
We can then subtract equation 2 from equation 1 and this gives us:
7x - 21 = 0
7x = 21
x = 3
We can then plug this back in and solve for y using either equation.

Veritas Help

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 Post subject: Re: Book 2 - Numeber 18 pg 31Posted: Tue Dec 06, 2011 5:26 pm

Joined: Fri Mar 18, 2011 12:27 pm
Posts: 17
Thank you both! That helped.

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