Solve for C: S = C + rC
- Thread starter kidd187
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mmm4444bot
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kidd187 said:
They told you what to do first. I'll repeat the same instruction, using different words:
Factor out the number C from the expression C + rC.
Here are some examples of this type of factoring.
Factor out the number A from the exression A + Ax:
A(1 + x)
Factor out the number k from the expression k + kM:
k(1 + M)
Factor Z as the GCF in the expression SZ - Z:
Z(S - 1)
We can see that these factorizations are correct because the Distributive Property shows us how to multiply the factors together, and, after we multiply them together, we get back the original expression.
A(1 + x) = (A)(1) + (A)(x) = A + Ax
k(1 + M) = (k)(1) + (k)(M) = k + kM
Z(S - 1) = (Z)(S) - (Z)(1) = SZ - Z
Let me repeat this.
A(1 + x) is the factored form.
A + Ax is the expanded form. (The word "expanded" means the multiplication was done.)
If you're given the factored form A(1 + x), then the Distributive Property gives you the expanded form A + Ax.
If you're given the expanded form A + Ax, then reversing the Distributive Property gives you back the factored form A(1 + x).
In your exercise, the expression on the right-hand side of the equation is written in expanded form C + rC.
The first step is to factor the expression C + rC.
Try it, and show me what you get.
After I see that you understand this first step, then we'll go to the next step toward solving the equation for C.
By the way, if you do not know what the phrase "solve for C" means, it means to rewrite the equation so that there is only a single symbol C, and this symbol is all by itself on one side of the equals sign:
C = something