3 basic algebra problems!!!!! [[please help me solve them read DEATILS!]]?

Question

x-c / c+a = a how do you solve for 'x' in this problem?

q(x+1)=s how do you solve for 'x'?

cx+a=bx how do you solve for 'x'?

where i put how do you solve for 'x' in this problem, it kind of mixed with my problem variables **keep that in mind!!!!!!!!!!!!!!~**

thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

x-c / c+a = a <=> x -a = a(c+a) <=> x = a(c+a+1)
q(x+1)=s <=> qx = s - q <=> x = (s-q)/q
cx+a=bx <=> x(c-b) = -a <=> x = a/(b-c)

Answer 2

x-c/c+a =a multiply both sides by (c+a) to get

a(c+a)= x-c then add c to both sides to get

a(c+a)+c =x very simple algebra, keep getting x term by itself
-----------------
q(x+1} = s multiply both sides by q to get

x+1 = sq sbutract 1 both sides to get

x= sq-1 apparently you do nto understand that in a equation
both sides can have any quantity added to it or
subtracted from it or both sides can be multiplied by
the same number (ALL without changing the
equality of the expression) It is as if the equal sign
was the midpoint of a weight scale with each side
of the equation balanced. On a scale you could add
the same amount to both sides or subtract the
same amount from both sides or multiply both sides
by the same amount and the scale would stay in
, balance.

So on to the third cx+a = bx This time get all the x terms on hte same side

(1) subtract cx from both sides to get

a = bx - cx then factor the x giving

a = x(b-c) then simply divide both sides by (b-c) to get

a/ (b-c) = x. One big word of caution , although you can add , subtract, aand multiply both sides of an equation by any mumber and keep the equality the same, you cannot divide by
zero, so in the last answer you would have to state that IF b=c.
then there is no answer to the equation for x

Answer 3

1. First: get rid of fractions > take the denominator (c + a) and multiply it by everything on both sides:

(c + a)[(x - c)/(c + a)] = (c + a)(a)

Second: cancel "like" terms:

x - c = (c + a)(a)
x - c = a(c) + a(a)
x - c = ac + a^2

Third: solve for "x" by keeping it on one side by adding "c" to both sides: opposite side means opposite sign

x - c + c = ac + a^2 - c
x = ac + a^2 - c

2. first: divide both sides by "q"

[q(x + 1)]/q = s/q

Second: cancel "like" terms:

x + 1 = s/q

Third: solve for "x" by keeping it on one side by subtracting 1 from both sides:

x + 1 - 1 = (s/q) -1
x = (s/q) -1

3. First: subtract "bx" from both sides and equal the equation to zero:

cx + a - bx = bx - bx
cx + a - bx = 0

Second: subtract "a" from both sides:

cx + a - a - bx = 0 - a
cx - bx = -a

Third: factor (cx - bx) by finding a common factor which is "x"

x(c - b) = - a

Fourth: solve for "x" by keeping it on one side > divide both sides by (c - b):

[x(c - b)]/c - b = -a/c - b

x = -a/c - b

Answer 4

Assume you meant this

(x - c)/(c + a) = a

x - c = a(a + c) {multiply both sides by (c + a)

x = a(a + c) + c {add c}

q(x + 1) = s
x + 1 = s/q {divide by q}
x = s/q - 1 {subtract 1}


cx + a = bx
cx - bx + a = 0 {subtract bx
cx - bx = - a {subtract a}
(c - b)x = - a {factor}
x - - a/(c - b) = a/(b - c)

Answer 5

we could basically say that the section= 30ft squared and the peak=5 A=a million/2bh (30)=a million/2b(5) first u can divide the two facets via a million/2 or 5 it quite is not substantial yet i ought to upload that when fixing for "x" you initiate with the backside of P.E.M.D.A.S (im useful you already know what it stands for) so what ever is subtracting (you will possibly upload to the different area) or including (you will possibly subtract from the different-area) it quite is not substantial then so on yet basically be useful you simplify on the area that has the "x" first or you will make it confusing for your self lol!! i think of im giving too lots element ......... (30)/(a million/2)=b(5) (15)=b(5) (15)/(5)=b B=3

Answer 6

x-c/(c+a)=a
multiply by (c+a)
x-c=a(c+a)
adding c
=c+a(c+a)

2.q(x+1)=s
dividing by q
x+1=s/q
ading -1
x=(s/q)-1

3.cx+a=bx
add -a-bx
cx-bx=-a
x(c-b)=-a
x(b-c)=a
divivding by b-c
x=a/(b-c)