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(-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

Sagot :

[tex]y=\frac{7x}{3}+32[/tex]

Explanation

Step 1

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

[tex]\begin{gathered} 7x-3y=4 \\ -3y=4-7x \\ y=\frac{-4}{3}+\frac{7x}{3}\rightarrow y=\text{ mx+b} \\ \\ \text{then} \\ \text{slope}=\frac{7}{3} \end{gathered}[/tex]

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

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

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