Answer: Anything larger than the fraction 21/127
21/127 = 0.16535 approximately
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Step-by-step explanation:
Let's solve each inequality for x
We'll start with the first inequality
![3(a-5x) < 1+x\\\\3a-15x < 1+x\\\\3a-1 < x+15x\\\\3a-1 < 16x\\\\16x > 3a-1\\\\x > (3a-1)/(16)](https://img.qammunity.org/2022/formulas/mathematics/high-school/kwszlpk9tqmfvsf8bbjbzi20jdvnkux373.png)
The variable x is larger than (3a-1)/16. If we knew what the value of 'a' was, then we could nail down the range of values for x more concretely.
Do the same for the second inequality
![2 - (x)/(2) > 3 + 5(x-a)\\\\2 - (1)/(2)x > 3 + 5(x-a)\\\\2 - 0.5x > 3 + 5(x-a)\\\\2 - 0.5x > 3 + 5x-5a\\\\2 + 5a-3 > 5x+0.5x\\\\5a-1 > 5.5x\\\\5.5x < 5a-1\\\\x < (5a-1)/(5.5)\\\\](https://img.qammunity.org/2022/formulas/mathematics/high-school/m1cg3ner6bzs086wkxtzi3hen4zmi117fp.png)
The variable x is also less than (5a-1)/(5.5)
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At this point, we have found that
and
![x < (5a-1)/(5.5)](https://img.qammunity.org/2022/formulas/mathematics/high-school/lfgw7r7hbqsmw4g23j3ufkkz9wxrs0ig1h.png)
This means x is between those endpoints, excluding each endpoint
So we can form this compound inequality
![(3a-1)/(16) < x < (5a-1)/(5.5)](https://img.qammunity.org/2022/formulas/mathematics/high-school/z4tlfmbv8ok9t01gnsu8h76i24hi1l7ikm.png)
This tells us the range of where x spans from.
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If the endpoints are equal, aka the same value, then there's no way to have x have any solutions
So let's equate the endpoints and see what happens
![(3a-1)/(16) = (5a-1)/(5.5)\\\\5.5(3a-1) = 16(5a-1)\\\\16.5a - 5.5 = 80a - 16\\\\-5.5 + 16 = 80a - 16.5a\\\\10.5 = 63.5a\\\\63.5a = 10.5\\\\a = (10.5)/(63.5)\\\\a = (105)/(635)\\\\a = (21*5)/(127*5)\\\\a = (21)/(127)\\\\a \approx 0.16535\\\\](https://img.qammunity.org/2022/formulas/mathematics/high-school/7pq2bwmlyabc36rtqwxbgyf5e8x0r38gbu.png)
If 'a' is equal to that value, then the two endpoints of that compound inequality are completely identical; therefore, in that situation, we wouldn't have any solutions for x.
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The question is: what happens if 'a' is some number smaller than 21/127?
Let's say a = 0
- The left endpoint would be (3a-1)/16 = (3*0-1)/16 = -0.0625
- The right endpoint would be (5a-1)/(5.5) = (5*0-1)/(5.5) = -0.1818 approximately
So we see that -0.0625 < x < -0.1818; however, upon closer inspection, you should find that such an interval makes no sense. Why not? Because -0.0625 is not smaller than -0.1818. It should be the other way around. No number exists that is between -0.0625 and -0.1818 (I recommend using a number line to help see why this is the case).
Therefore, anything smaller than a = 21/127 will result in having no solutions for x in the original system of inequalities.
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Now let's try something larger than a = 21/127
Let's say we picked a = 1
- The left endpoint would be (3a-1)/16 = (3*1-1)/16 = 0.125
- The right endpoint would be (5a-1)/(5.5) = (5*1-1)/(5.5) = 0.7273 approximately
We end up with 0.125 < x < 0.7273 which is now valid because 0.125 is indeed smaller than 0.7273
Therefore, if a > 21/127, then we'll have those original system of inequalities lead to more than one solution for x.