Hi can someone help with this math question please
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Explanation from Alloprof
This Explanation was submitted by a member of the Alloprof team.
Hi 😊
You have to use inequalities to solve.
First, set your variables.
x : number of lamps that have an electrical defect
y : number of lamps that have an optical defect
z : number of lamps that have both defects
Then, make your inqualities.
200 - 180 = 20
20 lamps have at least one defect.
x < 16
y < 12
x + y + z = 20
You have to resolve :
If you substract one lamp of each variable (x and y), you add one to z.
Try with :
x = 15
y = 12
z = 1
You have :
15 + 12 + 1 = 28 ≠ 20
You have to continue.
x = 10
y = 8
z = 6
10 + 8 + 6 = 24 ≠ 20
Again!
x = 6
y = 4
z = 10
6 + 4 + 10 = 20
We did it! 😎
So, we have :
x = 6 lamps that have an electrical defect
y = 4 lamps that have an optical defect
z = 10 lamps that have both defects
Hope this helped! 😁 Read you soon on the Help Zone 🥴
Explanation from Alloprof
This Explanation was submitted by a member of the Alloprof team.
Hi 😊
You have to use inequalities to solve.
First, set your variables.
x : number of lamps that have an electrical defect
y : number of lamps that have an optical defect
z : number of lamps that have both defects
Then, make your inqualities.
200 - 180 = 20
20 lamps have at least one defect.
x < 16
y < 12
x + y + z = 20
You have to resolve :
If you substract one lamp of each variable (x and y), you add one to z.
Try with :
x = 15
y = 12
z = 1
You have :
x + y + z = 20
15 + 12 + 1 = 28 ≠ 20
You have to continue.
x = 10
y = 8
z = 6
x + y + z = 20
10 + 8 + 6 = 24 ≠ 20
Again!
x = 6
y = 4
z = 10
x + y + z = 20
6 + 4 + 10 = 20
We did it! 😎
So, we have :
x = 6 lamps that have an electrical defect
y = 4 lamps that have an optical defect
z = 10 lamps that have both defects
Hope this helped! 😁 Read you soon on the Help Zone 🥴