# Help with solving linear equation

• Apr 9th 2009, 06:21 PM
erj145aviator
Help with solving linear equation
Problem: 1/3 + 1/9(k+5) - k/4 = 2

Question... When you multiply by 36 (LCD) to kill the fractions...

Why do you not multiply (k+5)?

I mean 36(1/3 + 1/9(k+5) - k/4) ... (k+5) is there... if you distribute 36 over... (k+5) would be affected. So why in the process of solving, would you not do that?
• Apr 9th 2009, 06:43 PM
x y z
Ok... when you multiply by 36, you multiply BOTH sides. It doesn't matter whether (x+5) would be affected or not, just as long as you multiply both sides by 36. Therefore, 1/3 + 1/9(k+5) - k/4 = 2 would become 12+4(k+5)-9k = 72, and then you would expand.
• Apr 9th 2009, 07:15 PM
mollymcf2009
Quote:

Originally Posted by erj145aviator
Problem: 1/3 + 1/9(k+5) - k/4 = 2

Question... When you multiply by 36 (LCD) to kill the fractions...

Why do you not multiply (k+5)?

I mean 36(1/3 + 1/9(k+5) - k/4) ... (k+5) is there... if you distribute 36 over... (k+5) would be affected. So why in the process of solving, would you not do that?

$\frac{1}{3} + \frac{1}{9}(k+5) - \frac{k}{4} = 2$

$\frac{1}{3} + \frac{(k+5)}{9} - \frac{k}{4} = 2$

$\frac{12}{36} + \frac{4(k+5)}{36} - \frac{9k}{36} = 2$

$12 + 4(k+5) - 9k = 72$

$12 + 4k + 20 - 9k = 72$

$-5k + 32 = 72$

$-5k = 40$

$k = -8$
• Apr 10th 2009, 12:02 PM
Marlin
Quote:

Originally Posted by erj145aviator
Problem: 1/3 + 1/9(k+5) - k/4 = 2

Question... When you multiply by 36 (LCD) to kill the fractions...

Why do you not multiply (k+5)?

I mean 36(1/3 + 1/9(k+5) - k/4) ... (k+5) is there... if you distribute 36 over... (k+5) would be affected. So why in the process of solving, would you not do that?

As far as I know you don't multiply k+5 by 35 because of the distributive property. You are keeping proportion with the rest of the equation.