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Sagot :
Consider that the slope-intercept form of the straight line with slope (m) and y-intercept (c) is given by,
[tex]y=mx+c[/tex]a.
Modify the given equation as,
[tex]\begin{gathered} \frac{x}{3}+\frac{y}{2}=1 \\ \frac{y}{2}=-\frac{x}{3}+1 \\ y=-\frac{2}{3}x+2 \end{gathered}[/tex]Thus, the equation in slope-intercept form can be written as,
[tex]y=-\frac{2}{3}x+2[/tex]b.
Modify the given equation as,
[tex]\begin{gathered} 4x-3y+2=0 \\ 3y=4x+2 \\ y=\frac{4}{3}x+\frac{2}{3} \end{gathered}[/tex]Thus, the equation in slope-intercept form can be written as,
[tex]y=\frac{4}{3}x+\frac{2}{3}[/tex]c.
Modify the given equation as,
[tex]\begin{gathered} x-y=5(x-y) \\ x-y=5x-5y \\ 5y-y=5x-x \\ 4y=4x \\ y=x \end{gathered}[/tex]Thus, the equation in slope-intercept form can be written as,
[tex]y=x[/tex]
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