Write an equation of the line that passes through (2,−5) and is parallel to the line 2y=3x+10
1 answer:
Step-by-step explanation:
Divide two on both sides to get rid of it and the make the equation in the form y = mx + c
y = 3/2x + 5
since both lines are parallel they must have the same gradient which is 3/2
y = 3/2x + c
All you have to do now is to replace x and y with (2, -5) to find c
x = 2
y = -5
-5 = 3/2 × 2 + c
-5 = 3 + c
c = -5-3
c = -8
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As a matter of simplification, same/same = 1.
now, "same" could be anything whatsoever, you name it, a whole polynomial, a whole anything whatsoever.
![\bf \cfrac{\sqrt{2}}{\sqrt{2}}=\cfrac{\sqrt{3}}{\sqrt{3}}=\cfrac{100000000}{100000000}=\cfrac{eggs}{eggs}=\cfrac{miles}{miles}=\cfrac{\frac{\quad \sqrt{3}}{\sqrt[5]{17}}\quad }{\frac{\sqrt{3}}{\sqrt[5]{17}}}=\cfrac{whatever}{whatever}=1](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B%5Csqrt%7B2%7D%7D%3D%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B3%7D%7D%3D%5Ccfrac%7B100000000%7D%7B100000000%7D%3D%5Ccfrac%7Beggs%7D%7Beggs%7D%3D%5Ccfrac%7Bmiles%7D%7Bmiles%7D%3D%5Ccfrac%7B%5Cfrac%7B%5Cquad%20%5Csqrt%7B3%7D%7D%7B%5Csqrt%5B5%5D%7B17%7D%7D%5Cquad%20%7D%7B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B%5Csqrt%5B5%5D%7B17%7D%7D%7D%3D%5Ccfrac%7Bwhatever%7D%7Bwhatever%7D%3D1)
now, that said, let us use 1 in this case then ... hmmmm we simply need a "y" in the denominator, so we simply multiply the denominator AND numerator by "y", y/y = 1 btw.
now, "same" could be anything whatsoever, you name it, a whole polynomial, a whole anything whatsoever.
now, that said, let us use 1 in this case then ... hmmmm we simply need a "y" in the denominator, so we simply multiply the denominator AND numerator by "y", y/y = 1 btw.