Inequalities

[ImagineDragons]

I haven't done these for at least a year, how do I do these again?

1. - 4x + 3y ≤ 15
2. - 6x – 4y < 12
3. -7x + 3y ≤ 15
4. 6x-5y < 12

[jim360]

Is the question about plotting these on a graph?

[ImagineDragons]

Nope it's just solving these

[vetuste_ennemi]

Are they supposed to have unique solutions?

[ImagineDragons]

Nope just the basic solutions

[vetuste_ennemi]

Despite my (surprisingly good) pass at O-Level Maths I'm pretty much illiterate in the discipline. However - taking 4 as an example - my intuition is that there is an infinite number of x/y values which meet the the requirement less than than 12.

[vetuste_ennemi]

Intuition is a fickle mistress. Maybe I should defer to the expert.

[JJ109]

I think that you are meant to describe x in terms of y or vice versa
3. -7x + 3y ≤ 15
Add 7x to both sides (like an equation)
3y ≤ 15 + 7x
Divide both sides by 3
y ≤ 5 + (7/3)x

[JJ109]

^^^ I haven't done these for 45 years, so I am a bit rusty also

[Jackdaw33]

That hardly gives you the values of x and y. All I can remember is that with two unknowns you need two simultaneous equations to solve it.

[JJ109]

They are inequalities - they are not meaant to be solved
One other point - when multiplying and dividing by a negative number, remember to flip the sign - so that greater than becomes less than and vice versa

[JJ109]

Therefore if you wanted to express number three as x in terms of y
-7x + 3y ≤ 15
subtract 3y from both sides
-7x ≤ 15 -3y
divide both sides by -7
x >= (greater and equal) (15 - 3y) / 7

[davegosling]

"4. 6x-5y < 12"

x=5
y=4

6x5 = 30, 5x4 = 20
30 - 20 = 10 and 10 < 12

[Jackdaw33]

You could just as well say x = 10 and y =11 to achieve the same result.

[JJ109]

@davegosling tht is only one solution but there are an infinite number
4
x = 2, y = 3
x = 2, y = 4
x = 20, y = 30
etc etc

[davegosling]

Indeed. Each has many possible answers.

[Prudie]

There are infinite solutions if x and y can be negatives or fractions. The best you can do is plot a straight line graph showing these infinite solutions.

[JJ109]

I think all you need to do is express x in terms of y. Therefore,
1. -4x + 3y ≤ 15
Subtract 3y from both sides
-4x ≤ 15 -3y
Divide by - 4 (flip the inequality sign)
x >=(greater than or equal) (15 -3y)/4

2. - 6x – 4y < 12
-6x < 12 + 4y
Divide by -6 (flip the inequality sign)
x > (12 + 4y)/6
x > (6 + 2y)/ 3

3. already done

4. 6x - 5y < 12
6x < 2 + 5y
divide by 6 (as positive do NOT flip the sign)
x < (2 + 5y) / 6

[JJ109]

ERRORS in my earlier posts as I forgot to divide the right hand side by minus also. I told you I was rusty
-4x + 3y ≤ 15
-4x ≤ 15 - 3y
x >= -(15-3y)/7
x>= (3y - 15) / 7

2. -6x - 4y < 12
-6x < 12 + 4y
x > -(12 + 4y) / 6
x > -(6 + 2y) / 3
x > -(2/3)(3 + y)

3. -7x + 3y ≤ 15
-7x ≤ 15 -3y
x >= - (5 -3y)/ 7

[fiction-factory]

Yes you can rewrite the inequalities -eg change 3 to y≤(7x/3) +5, but that doesn't take you much further forwards.
We need to know the exact wording of the question.
Maybe you are supposed to plot them all and shade the region (if any) which satisfies the four inequalities. I can't try it now as I'm off to work but I'll look back later if you can confirm the exact wording of the question.

[fiction-factory]

I think that's what prudie is suggesting