I came up with x = 2c + a squared. I'm not sure if this right though. I started working the problem by getting rid of the fraction by multiplying c+a with x-c to cross out the c+a on that side, then I multiplied c+a times a .... Here's where I'm not sure if I should multiply c+a times "a" using PEMDAS or if I should multiply "a" times c+a using the distributive property. Help.
2007-03-18 03:41:33 · 7 answers · asked by Toni 1 in Science & Mathematics ➔ Mathematics
(x-c)/(c+a) = a
x-c = a(c+a)
x = a(c+a) + c
use the distributive property
x = ac+a^2+c
2007-03-18 03:44:23 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I ended up with x = a^2 + ac + c
2007-03-18 10:47:03 · answer #2 · answered by Drools over home made food 6 · 0⤊ 0⤋
i got x=a(squared) +ac+c
by multiplying both sides by c+a and them taking across the c
2007-03-18 10:53:52 · answer #3 · answered by multiplayertim 2 · 0⤊ 0⤋
x = ac+a^2+c
2007-03-18 10:57:44 · answer #4 · answered by Anonymous · 0⤊ 0⤋
(x-c)/(c+a) = a
x- c = c(c+a)
= c^2 + ac (using distributive property)
so x = c^2 + ac - c
2007-03-18 10:45:24 · answer #5 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
(x-c)/(c+a) = a
x-c = a(c+a) multiply the denominator (c+ a) by a....
x-c = ac + a²
x= ac + a² + c
x= a² + ac + c ............ans.
2007-03-18 10:52:14 · answer #6 · answered by edison c d 4 · 0⤊ 0⤋
(x-c)/(c+a) = a
x-c = (c+a)a
x-c = ac + a**2
x = a**2 + ac + c
2007-03-18 10:46:28 · answer #7 · answered by rdbart3 1 · 0⤊ 0⤋