Question

(-15,-3);7x-3y=4 , find the equation of the line that passes through the point given and is parallel to the function given,give the equation is slope-intercept point

Answer 1

`y=(7x)/(3)+32`

Step-by-step explanation

Step 1

we need to find the slope, the, isolate y from the equation

`\begin{gathered} 7x-3y=4 \\ -3y=4-7x \\ y=(-4)/(3)+(7x)/(3)\rightarrow y=\text{ mx+b} \\ \\ \text{then} \\ \text{slope}=(7)/(3) \end{gathered}`

m1=m2, then, the slope of the line that we are looking for is 7/3

Step 1

find the equation using P1(-15,-3) and slope )7/3

`\begin{gathered} y-y_1=m(x-x_1) \\ \text{replace} \\ y-(-3)=(7)/(3)(x-(-15)) \\ y+3=(7x)/(3)+(105)/(3) \\ y=(7x)/(3)+(105)/(3)-3 \\ y=(7x)/(3)+32 \\ \end{gathered}`

I hope this helps you