# Thread: Help with solving linear equation

1. ## 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?

2. 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.

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

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

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

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

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

$-5k + 32 = 72$

$-5k = 40$

$k = -8$

4. 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.