Ok so we'll go ahead and solve for y first - we just need to get it alone on one side of the equal sign
Step 1: subtract 2x from each side
2x - 7y - 2x = 19 - 2x
This cancels out the 2x on the left, giving us
-7y = 19 - 2x
Step 2: divide both sides by -7
=
+ ![\frac{-2x}{-7}](https://tex.z-dn.net/?f=%5Cfrac%7B-2x%7D%7B-7%7D)
This gives us
y = -19/7 + 2x/7
That should be your answer for the first question. Now solving the next parts are easy. All you need to do is plug in x.
When x = -3
y = -19/7 + 2x/7
y = -19/7 + 2(-3)/7
y = -19/7 - 6/7
y = -25/7
When x = 0
y = -19/7 + 2x/7
y = -19/7 + 2(0)/7
y = -19/7
When x = 3
y = -19/7 + 2x/7
y = -19/7 + 2(3)/7
y = -19/7 + 6/7
y = -13/7
Hope that helps! Feel free to ask if I can help with anything else :)
When you see questions of this nature, test the individual inequalities and look out for their intersection.
For
![y < \frac{2}{3} x](https://tex.z-dn.net/?f=y%20%3C%20%5Cfrac%7B2%7D%7B3%7D%20x)
Choose a point in the lower or upper half plane created by the line
![y = \frac{2}{3} x](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20x)
The above line is the one which goes through the origin.
Now testing (1,0) yields,
![0 < \frac{2}{3} (1)](https://tex.z-dn.net/?f=0%20%3C%20%5Cfrac%7B2%7D%7B3%7D%20%281%29)
That is,
![0 < \frac{2}{3}](https://tex.z-dn.net/?f=0%20%3C%20%5Cfrac%7B2%7D%7B3%7D%20)
This statement is true. So we shade the lower half of
![y = \frac{2}{3} x](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20x)
For
![y \geqslant - x + 2](https://tex.z-dn.net/?f=y%20%5Cgeqslant%20-%20x%20%2B%202)
We test for the origin because, it is not passing through the origin.
![0 \geqslant - (0) + 2](https://tex.z-dn.net/?f=0%20%5Cgeqslant%20-%20%280%29%20%2B%202)
This yields
![0 \geqslant 2](https://tex.z-dn.net/?f=0%20%5Cgeqslant%202)
This statement is false so we shade the upper half.
The intersection is the region shaded in B. The top right graph
Let
![S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n](https://tex.z-dn.net/?f=S_n%20%3D%20%5Cdisplaystyle%20%5Csum_%7Bk%3D0%7D%5En%20r%5Ek%20%3D%201%20%2B%20r%20%2B%20r%5E2%20%2B%20%5Ccdots%20%2B%20r%5En)
where we assume |r| < 1. Multiplying on both sides by r gives
![r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}](https://tex.z-dn.net/?f=r%20S_n%20%3D%20%5Cdisplaystyle%20%5Csum_%7Bk%3D0%7D%5En%20r%5E%7Bk%2B1%7D%20%3D%20r%20%2B%20r%5E2%20%2B%20r%5E3%20%2B%20%5Ccdots%20%2B%20r%5E%7Bn%2B1%7D)
and subtracting this from
gives
![(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}](https://tex.z-dn.net/?f=%281%20-%20r%29%20S_n%20%3D%201%20-%20r%5E%7Bn%2B1%7D%20%5Cimplies%20S_n%20%3D%20%5Cdfrac%7B1%20-%20r%5E%7Bn%2B1%7D%7D%7B1%20-%20r%7D)
As n → ∞, the exponential term will converge to 0, and the partial sums
will converge to
![\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20S_n%20%3D%20%5Cdfrac1%7B1-r%7D)
Now, we're given
![a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a](https://tex.z-dn.net/?f=a%20%2B%20ar%20%2B%20ar%5E2%20%2B%20%5Ccdots%20%3D%2015%20%5Cimplies%201%20%2B%20r%20%2B%20r%5E2%20%2B%20%5Ccdots%20%3D%20%5Cdfrac%7B15%7Da)
![a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}](https://tex.z-dn.net/?f=a%5E2%20%2B%20a%5E2r%5E2%20%2B%20a%5E2r%5E4%20%2B%20%5Ccdots%20%3D%20150%20%5Cimplies%201%20%2B%20r%5E2%20%2B%20r%5E4%20%2B%20%5Ccdots%20%3D%20%5Cdfrac%7B150%7D%7Ba%5E2%7D)
We must have |r| < 1 since both sums converge, so
![\dfrac{15}a = \dfrac1{1-r}](https://tex.z-dn.net/?f=%5Cdfrac%7B15%7Da%20%3D%20%5Cdfrac1%7B1-r%7D)
![\dfrac{150}{a^2} = \dfrac1{1-r^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B150%7D%7Ba%5E2%7D%20%3D%20%5Cdfrac1%7B1-r%5E2%7D)
Solving for r by substitution, we have
![\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)](https://tex.z-dn.net/?f=%5Cdfrac%7B15%7Da%20%3D%20%5Cdfrac1%7B1-r%7D%20%5Cimplies%20a%20%3D%2015%281-r%29)
![\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B150%7D%7B225%281-r%29%5E2%7D%20%3D%20%5Cdfrac1%7B1-r%5E2%7D)
Recalling the difference of squares identity, we have
![\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}](https://tex.z-dn.net/?f=%5Cdfrac2%7B3%281-r%29%5E2%7D%20%3D%20%5Cdfrac1%7B%281-r%29%281%2Br%29%7D)
We've already confirmed r ≠ 1, so we can simplify this to
![\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15](https://tex.z-dn.net/?f=%5Cdfrac2%7B3%281-r%29%7D%20%3D%20%5Cdfrac1%7B1%2Br%7D%20%5Cimplies%20%5Cdfrac%7B1-r%7D%7B1%2Br%7D%20%3D%20%5Cdfrac23%20%5Cimplies%20r%20%3D%20%5Cdfrac15)
It follows that
![\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12](https://tex.z-dn.net/?f=%5Cdfrac%20a%7B1-r%7D%20%3D%20%5Cdfrac%20a%7B1-%5Cfrac15%7D%20%3D%2015%20%5Cimplies%20a%20%3D%2012)
and so the sum we want is
![ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}](https://tex.z-dn.net/?f=ar%5E3%20%2B%20ar%5E4%20%2B%20ar%5E6%20%2B%20%5Ccdots%20%3D%2015%20-%20a%20-%20ar%20-%20ar%5E2%20%3D%20%5Cboxed%7B%5Cdfrac3%7B25%7D%7D)
which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?
You would start out by putting 225 over 100 then simplifying from there. You would end up getting 9/4. As a mixed number that would be 2 and 1/4.