To solve the exercise you can use the point-slope formula, that is,
![\begin{gathered} $y-y_1=m(x-x_1)$ \\ \text{ Where m is the slope of the line and } \\ (x_1,y_1)\text{ is a point through which the line passes} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z4g5ndsr6nqnv43xci0ng3s896zdmcir67.png)
So, in this case, you have
![\begin{gathered} m=(3)/(4) \\ (x_1,y_1)=(3,7) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/j0siw5g5lriqsorktcczczv4zdigohpee0.png)
![\begin{gathered} y-y_1=m(x-x_1) \\ \text{ Replace} \\ y-7=(3)/(4)(x-3) \\ y-7=(3)/(4)x-3\cdot(3)/(4) \\ y-7=(3)/(4)x-(9)/(4) \\ \text{ Add 7 from both sides of the equation} \\ y-7+7=(3)/(4)x-(9)/(4)+7 \\ y=(3)/(4)x+(19)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cwovucpfqz9x1d1pzxk1bl69vya2aak6cx.png)
Therefore, the equation of the line is
![y=(3)/(4)x+(19)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/a5u77m44amwxufurkxwrvci7xo8x5w9c7l.png)
Finally, to find the x-coordinate of point A, replace y = 5 into the equation of the line you just found and solve for x
![\begin{gathered} y=(3)/(4)x+(19)/(4) \\ 5=(3)/(4)x+(19)/(4) \\ \text{ Subtract 19/4 from both sides of the equation} \\ 5-(19)/(4)=(3)/(4)x+(19)/(4)-(19)/(4) \\ (1)/(4)=(3)/(4)x \\ \text{ Multiply by 4/3 on both sides of the equation} \\ (1)/(4)\cdot(4)/(3)=(4)/(3)\cdot(3)/(4)x \\ (1)/(3)=x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3g8m6zrvstz6wkp58pnh891vjvt0y66rj4.png)
Therefore, if point A(x,5) lies on the line, the value of x is 1/3.