$3(2x+d)+c(x+5)=10x+17$

Question

Given "$3(2x+d)+c(x+5)=10x+17$" what are the values of c and d. Upon expansion we get $6x+3d+cx+5c=10x+17$, meaning c must be equal to 4.

I was playing around with the equation and I found something I can't quite understand.

$$6x+3d+cx+5c=10x+17,$$

$$6x+cx=10x+17-3d-5c,$$

$$cx-4x=17-3d-5c$$

$$x=\frac{17-3d-5c}{c-4}$$

Previously I found $c$ to be $4$, but plugging in the value of c into the above, gives the denominator a value of zero, which is undefined upon division. What does this mean? Does a real value of x not exist or is x undefined. Could someone please help me to understand what is going on?

Answer 1

If you sub $c=4$ into the expanded equation you can quickly find that $d=-1$. If you sub both these values into your final expression for $x$ you get $x=\frac{0}{0}$. In other words $x$ could be anything.

The problem is that you are in effect trying to find $x$ by solving the equation $10x+17=10x+17$ which of course cannot give you a specific result because it is true for all $x$

Answer 2

Maybe a more general remark: Your eqution has three unknowns ($c$, $d$, and $x$), so you have to specify what you want to solve for.

The naming of the unknown suggests that you consider the left and right-hand side of your equation as functions of $x$, where the left hand-side depends on two parameters $c$ and $d$, and you want them to be equal. Thus, you're not lookign for a specific value of $x$ such that the equation holds, but you want it to be true for all $x$.

Thus, once you can calculate the parameters as in robert's answer, the two sides are identical (by construction), and you cannot also fix a value of $x$.