The relationship between t and r is expressed by the equation 2t + 3r + 6= 0. If r increases by 4

Question

The relationship between $t$ and $r$ is expressed by $$2t + 3r+ 6 = 0$$

If $r$ increases by $4$, what change would occur in $t$.

How do I solve this problem step by step? Also, what is the name of this kind of math problem so I can look up for reference and practice problems?

Answer 1

Well (Here $t_2$ is new $t$ after increasing $r$ by 4

Since$$2t+3r+6=0$$ $$2t_2+3(r+4)+6=0$$ $$2t_2+3r+12+6=0$$ Substituting $3r+6=-2t$ $$2t_2+12-2t=0$$ SO $$t_2=t-6$$

Answer 2

We have $$2t+3r+6=0$$ $$\iff t=-\left(\frac{3r+6}{2}\right) =\color{blue}{-\frac{3}{2}r-3}$$

So we now have $t$ expressed as a function of $r$. In fact, when graphed, we can see that the equation is that of a line:

So if any given value of $r$ changes, so will will the value $t$.

To give you a general idea of the message immediately above, that when $r= -2$, $t = -\frac 32(-2) - 3 = 3-3= 0.$ And we can see that the line intersects the r-axis at the point $(r, t) = (-2, 0)$. When $r = -2 + 4 = 2$, $t= -\frac 32(2) - 3 = -6$.

To see why this happens, whatever the value of the original $r$, we can find $t^*$ (the new value of $t$) when we replace $r$ by $r^*= r+4$.

$$\begin{align} t^* &= -\frac 32(r^*)- 3 \\ \\ &= -\frac 32(r+4) - 3 \\ \\ &= -\frac 32r- 6-3 \\ \\ &= \color{blue}{\left(-\frac 32r - 3\right)}- 6\\ \\ &= t - 6\end{align}$$

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So, when $r$ increases by $4$, $t$ decreases by 6.