# Give an equation for c not involving a.?

Combine c=4a-3 and 2b=3a-8

Relevance
• c=4a-3--------(1)

2b=3a-8--------(2)

3(1)-4(2)

=>

3c-8b=-9+32

=>

3c=23+8b

=>

c=(23+8b)/3

• c=4a-3...(1), 2b=3a-8...(2).;

(2)---> a = (1/3)2(b+4) = (2/3)(b+4).;

Then (1)---> c = 4[(2/3)(b+4)] -3 = (8/3)b +(32/3) -(96/3) = (8/3)b -(64/3) = (8/3)(b-8);

ie., c = (8/3)(b-8).

• An equation for c not involving a.

Combine c = 4a - 3 and 2b = 3a - 8

a = (c + 3)/4

• 12a = 3(c + 3) = 4(2b + 8)

3c = 8b + 23

• 1)

c = 4a - 3 ===> c + 3 = 4a ===> a = (c + 3)/4.

2)

2b = 3a - 8 ===> 2b + 8 = 3a ===> a = (2b + 8)/3.

3)

If "a = (c + 3)/4" and "a = (2b + 8)/3", therefore:

(c + 3)/4 = (2b + 8)/3.

3 * (c + 3) = 4 * (2b + 8).

3 * c + 3 * 3 = 4 * 2b + 4 * 8.

3c + 9 = 8b + 32.

3c = 8b + 32 - 9.

3c = 8b + 23.

c = (8b + 23)/3 <=== ANSWER.

Or, alternatively: c = (8/3) * b + 23/3.

• c = 4a - 3 and 2b = 3a - 8

The second equation can be re-arranged to make a the subject

so, 3a = 2b + 8

Hence, a = (2b + 8)/3

Putting this into the first equation for a gives:

c = 4(2b + 8)/3  - 3

or, c = 8b/3 + 32/3 - 9/3

so, c = 8b/3 + 23/3

Then, c = (8b + 23)/3....which gives c in terms of b

:)>

• given c=4a-3, now 4a=3-c, => a=3-c/4  & similarly a=2b+8/3 , solve them

• If you want to remove the "a", then solve one equation for "a" in terms of the other variable and substitute into the other equation.  This will give you a new equation with two unknowns, "b" and "c".  Then you can solve for c.

So we have:

c = 4a - 3 and 2b = 3a - 8

The first is already c in terms of a, so we want to leave that alone for now.  Solve the other one for a in terms of b:

2b = 3a - 8

2b + 8 = 3a

(2b + 8) / 3 = a

Now we can substitue that expression for "a" in the other equation:

c = 4a - 3

c = 4(2b + 8) / 3 - 3

Let's distribute the 4 and get a common denominator so we can simplify this:

c = (8b + 32) / 3 - 9 / 3

Now we can subtract the numerators:

c = (8b + 32 - 9) / 3

and simplify:

c = (8b + 23) / 3

or:

c = (8/3)b + 23/3

Either gives you c in terms of b.