True or False? The following is a system if Linear inequalities x(2-y)>1 x+3y less than or = to 4

Answer 1

True.

It is a system of linear inequalities.
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Answer 2

I assume that ur asking: is this system possible? Which is true as i will explain now, I assume some prior knowledge of solving quadratic equations etc. if u don't understand something give me a shout.

Lets lable the following equations as follows:
x(2-y)>1 equation (1)
x+3y<=4 equation (2)

In mathematics the best strategy to solve inequalities is to first solve the equations as if both have an equals signs as shown:
x(2-y)=1 (1) I'll ommit the word 'equation' from now on
x+3y= 4 (2)

Now to solve this equation we have to eliminate one of the unknowns, i have decided x, but it doesn't matter really.

rearranging (2)
x=4-3y
substituting into (1)

(4-3y)(2-y)=1

multiplying this out
8-4y-6y+3y^2 (3y squared)
3y^2-10y+7=0
Solving this quadratic equation
y=2 1/3 or 1 (quadratics have 2 roots)

now we remember that initially equation (1) had a > sign. hence 3y^2-10y+7>0 as we stuck (2) into (1)

Sticking in random numbers less than one, between 1 and 2 1/3 and more than 2 1/3, we find that this function is >0 when y<1 and 2 1/3 range needed for the quadratic inequality. we know that it's always >0 because the function can only change sign from before and then after the root.

lets choose any random number that is either <1 or>2 1/3. I chose 0 for simplicity. Putting into equation (1)
2x>1
x>1/2
Choose now a value for x greater than 1/2 e.g. 1
Sticking x=1 y=0 into (2)
1<=4 so those equations are true and solved for you:
x>0.5 y<1 or y>2 1/3
fiddle around with those numbers as much as you like provided x and y meet the above conditions, the inequalities work!!!

Answer 3

False, when you multiply out that first inequality to get 2x-xy > 1 you have a term xy which will is not linear

Answer 4

False x(2_y)>1x+3y = 2points thankyou.