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Question 481190: Can you please help me solve this I really need help I have tried many times but I get stuck and i need to give it in very soon.
-What is the solution to the following system of inequalities?
2x > -5 + 3y
2x − 5y < -6
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! solve for y in both equations to get:
y < (2x+5)/3
y > (2x+6)/5
set these equations to equality to get:
y = (2x+5)/3
y = (2x+6)/5
since they are both equal to y, then they are both equal to each other and you get:
(2x+5)/3 = (2x+6)/5
cross multiply to get:
3*(2x+6) = 5*(2x+5)
multiply out to get:
6x + 18 = 10x + 25
subtract 6x from both sides of the equation and subtract 18 from both sides of the equation to get:
-7 = 4x
divide both sides of the equation by 4 to get:
x = -7/4
when x = -7/4, both equations are equal to each other.
look at what happens when x is smaller than -7/4 and when x is > -7/4
i chose -2 for smaller and -1 for greater.
my results from solving for these values of x is shown in the following table.
x y1 = (2x+5)/3 y2 = (2x+6)/5
-7/4 .5 .5
-1 1 .8
-2 .33333333 .4
from this table you can see that:
when x is equal to -7/4, y1 = y2
when x is greater than -7/4, y1 is greater than y2.
when x is smaller than -7/4, y1 is smaller than y2.
we were looking to solve for:
y < (2x+5)/3
y > (2x+6)/5
since y1 = (2x+5)/3 and y2 = (2x+6)/5, then we are looking for a value of y that is smaller than y1 and greater than y2.
this is possible when y1 is greater than y2 which occurs when x is greater than -7/4.
the equation is satisfied when x > -7/4.
if we graph both the equations of:
y = (2x+5)/3
y = (2x+6)/5
we will see that the region of compatibility is when x > -7/4 as shown below.
the graph that crosses the y-axis closer to 1 is the equation of y = (2x+5)/3.
that becomes the lower line after the crossing point of x = -7/4.
you get:
the higher line is y = (2x+6)/5
the lower line is y = (2x+5)/3
any value of y between these lines will simultaneously satisfy the equations of:
y < (2x+6)/5
y > (2x+5)/3
this only occurs when x is greater than -7/4.
since that's what we originally set out to find, then the solution is good.
our answer is that the value of x that satisfy both requirements occurs when x > -7/4.
it helps to find the crossover point first.
once you do that, you can test for values of x above and below that value.
since these are linear equations, you will only find one crossover point.
when you get to higher order equations, the crossover points can be more than 1.
you will then get regions where the requirements of the simultaneous equations are satisfied.
for this set of equations, there is only one region that satisfies the requirements and that is when the value of x is greater than -7/4.
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