Respuesta :
Conider the equation given below;
[tex]6x+2y=13[/tex]To express this equation in the form
[tex]y=mx+b[/tex]We shall begin by making y the subject of the equation, a follows;
[tex]\begin{gathered} 6x+2y=13 \\ \text{Subtract 6x from both sides;} \\ 6x-6x+2y=13-6x \\ 2y=13-6x \\ \text{Divide both sides by 2, and you'll have;} \\ \frac{2y}{2}=\frac{13-6x}{2} \\ y=\frac{13-6x}{2} \\ \text{Simplify the right side;} \\ y=\frac{13}{2}-\frac{6x}{2} \\ y=\frac{13}{2}-3x \\ \text{Written in the slope-intercept form, it now becomes;} \\ y=-3x+\frac{13}{2} \end{gathered}[/tex]The values of m and b are;
[tex]\begin{gathered} y=mx+b \\ y=-3x+\frac{13}{2} \\ \text{Hence;} \\ m=-3,b=\frac{13}{2} \end{gathered}[/tex]Part B:
Therefore, for a point on this line where x = 2, we would have;
[tex]\begin{gathered} y=-3x+\frac{13}{2} \\ \text{Substitute for the value of x}=2 \\ y=-3(2)+\frac{13}{2} \\ y=-6+\frac{13}{2} \\ y=\frac{13}{2}-6 \\ \text{Take the LCM of both numbers and we'll now have;} \\ y=\frac{13-12}{2} \\ y=\frac{1}{2} \\ \text{Therefore the ordered pair would be;} \\ (2,\frac{1}{2}) \end{gathered}[/tex]ANSWER:
Part A;
[tex]y=-3x+\frac{13}{2}[/tex]Part B:
[tex](2,\frac{1}{2})[/tex]