Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$?

Question

Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$?

If $h(a)=a$, then $4x-1=2x+7$ which implies $x=4$. So $a=15$ when I substitute $x=4$ into both linear equations. Is the value of $x$ $15$?

Answer 1

The posted proof looks good. For an alternative one, let $4x-1=y \iff x = \frac{y+1}{4}\,$, then:

$$\,h(y) = h(4x-1)=2x+7=2\,\frac{y+1}{4}+7=\frac{y}{2}+\frac{15}{2}\,$$

Therefore $\,h(x)=\frac{x}{2}+\frac{15}{2}\,$, and $h(x)=x \iff x = \frac{x}{2}+\frac{15}{2} \iff x=15\,$.

Answer 2

First, we find an expression for $h(x)$. From our definition of $h$, we have $h(4y-1) = 2y+7$. So, if we let $x=4y-1$, so that $y = (x+1)/4$, we have[h(x) = 2\cdot\frac{x+1}{4} + 7 = \frac{x+1}{2} + 7.]Setting this equal to $x$ gives[x =\frac{x+1}{2} + 7.]Multiplying both sides by 2 gives $2x = x+1 + 14$, so $x = \boxed{15}$.