Solution:
Given the table as shown below:
Using the equation:
![y=mx+b\text{ ---- equation 1}](https://img.qammunity.org/2023/formulas/mathematics/college/g1q7bxfg89km9y2x58ojnvf4th54s66i7v.png)
When x = 68, y equals 4.1.
Thus, substitute these values into equation 1
![\begin{gathered} 4.1=m(68)+b \\ \Rightarrow68m+b=4.1\text{ ----- equation 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ajk4wr612nn7ia77633n90nx1vpviqywpt.png)
When x = 71, y equals 4.6.
Similarly, we have
![\begin{gathered} 4.6=m(71)+b \\ \Rightarrow71m+b=4.6\text{ ----- equation 3} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o3timsc0fvzcv9sin6s3gqkq4ntwwt0i21.png)
From equation 2, make c the subject of the formula.
![\begin{gathered} 68m+c=4.1\text{ } \\ \Rightarrow c=4.1-68m\text{ ---- equation 4} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oku56wuglxynx6xz55xt9q17e9z1z9ojkq.png)
Substitute equation 4 into equation 3. Thus,
![\begin{gathered} 71m+c=4.6\text{ } \\ \Rightarrow71m+(4.1-68m)=4.6 \\ \text{open parentheses} \\ 71m+4.1-68m=4.6 \\ \text{collect lik terms} \\ 71m-68m=4.6-4.1 \\ 3m=0.5 \\ \text{divide both sides by the coefficient of m, which is 3.} \\ \text{thus,} \\ m=(0.5)/(3) \\ \Rightarrow m=(1)/(6) \end{gathered}]()
Substitute the obtained value of m into equation 4.
thus,
![\begin{gathered} c=4.1-68m \\ \text{where m=}(1)/(6) \\ \text{thus,} \\ c=4.1-68((1)/(6)) \\ =4.1-11.33 \\ \Rightarrow c=-7.23 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9fa7tp0lk2tr7kzyuiped0gxa5jkjs8wge.png)
Hence, (x,y) Notation y=mx+c
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