Answer:
[tex]\textsf{Slope-intercept form}: \quad y=-\dfrac{4}{3}x+\dfrac{5}{3}[/tex]
[tex]\textsf{Standard form}: \quad 4x+3y=5[/tex]
Step-by-step explanation:
Given linear equation:
[tex]y-11=-\dfrac{4}{3}(x+7)[/tex]
The given linear equation is point-slope form.
[tex]\boxed{\begin{minipage}{3.7cm}\underline{Slope-intercept form}\\\\$y=mx+b$\\\\where $m$ is the slope\\ and $b$ is the $y$-intercept.\\\end{minipage}}[/tex]
To write the equation in slope-intercept form, isolate y:
[tex]\implies y-11=-\dfrac{4}{3}(x+7)[/tex]
[tex]\implies y-11=-\dfrac{4}{3}x-\dfrac{28}{3}[/tex]
[tex]\implies y-11+11=-\dfrac{4}{3}x-\dfrac{28}{3}+11[/tex]
[tex]\implies y=-\dfrac{4}{3}x+\dfrac{5}{3}[/tex]
[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Standard form}\\\\$Ax+By=C$\\\\where $A, B$ and $C$ are constants and $A$ must be positive.\\\end{minipage}}[/tex]
To write the equation in standard form, eliminate the fraction, bring the terms in x and y to the left of the equation, and the constants to the right:
[tex]\implies y-11=-\dfrac{4}{3}(x+7)[/tex]
[tex]\implies (y-11) \cdot 3=-\dfrac{4}{3}(x+7) \cdot 3[/tex]
[tex]\implies 3y-33=-4(x+7)[/tex]
[tex]\implies 3y-33=-4x-28[/tex]
[tex]\implies 3y-33+4x=-4x-28+4x[/tex]
[tex]\implies 3y-33+4x=-28[/tex]
[tex]\implies 3y-33+4x+33=-28+33[/tex]
[tex]\implies 4x+3y=5[/tex]
