The general equation of a line is expressed as
[tex]\begin{gathered} y\text{ = mx+c ----- equation 1} \\ \text{where} \\ m\Rightarrow slope\text{ of the line} \\ c\Rightarrow y-intercept\text{ of the line} \end{gathered}[/tex]When the linear function crosses the x-axis at x = -3, the value of y equals zero.
Thus, substitute the above parameters in equation 1.
[tex]\begin{gathered} 0\text{ = m(-3) + c} \\ \Rightarrow-3m\text{ + c = 0 ----- equation 2} \end{gathered}[/tex]When the linear function crosses the y-axis at y = 5, the value of x equals zero.
Thus, substitute the above parameters in equation 1.
[tex]\begin{gathered} 5\text{ = m(0) + c} \\ \Rightarrow c\text{ = 5 ------ equation 3} \end{gathered}[/tex]From equation 3, substitute the value of 5 for c in equation 2.
Thus,
[tex]\begin{gathered} -3m\text{ + c =0} \\ \Rightarrow-3m\text{ + 5 }=\text{ 0} \end{gathered}[/tex]solve for m,
[tex]\begin{gathered} -3m\text{ + 5 }=\text{ 0} \\ collect\text{ like terms} \\ -3m\text{ = 0-5} \\ -3m\text{ = -5} \\ \text{divide both sides of the equation by the coefficient of m, which is -3} \\ \frac{-3m}{-3}\text{ = }\frac{\text{-5}}{-3} \\ \Rightarrow m\text{ = }\frac{5}{3} \end{gathered}[/tex]Since the values of m and c are now known, substitute their respective values into
equation 1.
From equation 1, we have
[tex]\begin{gathered} y\text{ = mx + c} \\ \Rightarrow\text{ y = }\frac{5}{3}x\text{ + 5} \end{gathered}[/tex]Hence, the equation of the line is
[tex]\text{ y = }\frac{5}{3}x\text{ + 5}[/tex]