OK to solve this, we have to solve each system presented through elimination or substitution and find which one is equivalent to that of the teacher's!
First let's solve for the teacher's:
-2x+5y=10
-3x+9y=6
Solve by substitution (I think elimination might be easier to do for this one, but I don't really remember 100% sorry!)
Isolate the x (or y) variable in the first equation
-2x+5y=10
-2x=10-5y

Substitute x into the next equation and solve for y
-3(10-5y/2)+9y=6
3*10-5y/2+9y=6
(multiply both sides by 2)
3(10-5y)+18y=12
30-15y+18y=12
30+3y=12
3y=-18
y=-6
Substitute in x
x= -10-5(-6)/2
x=-20
TEACHER'S ANSWER (-20,-6)
GOKU
x-3y=-2
-2x+5y=-7
Do the same as above
Solve for x
x-3y=-2
x=3y-2
Plug in
-2(3y-2)+5y=-7
4-6y+5y=-7
4-y=-7
-y=-11
y=11
x=(3(11)-2)
x=31
GOKU'S ANSWER (31, 11)
SELINA:
-5x+14y=16
-3x+9y=12
One last time!! :)
-5x+14y=16
-5x=16-14y
x=(16-14y)/-5
-3(-(16-14y/5)+9y=12
3*16-14y/5+9y=12
3*16-14y+45y=60
48-42y+45y=60
48+3y=60
3y=12
y=4
x=-(16-14(4))/5
x=8
SELINA'S ANSWER
(8,4)
So neither Goku or Selina got the same answer as the teacher
Answer:
here go the answer hope it help
Step-by-step explanation:
I think this would be 4≥X<span>>-8
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Answer:
Step-by-step explanation:
On the day of the canteen, 800 coupons were sold, the price of each coupon was RM 30 and RM 50 respectively. The amount of money earned from the sale of coupons was RM30000. How many copies of RM30 and RM50 coupons were sold?
Let:
RM 30 = x
RM 50 = y
x + y = 800 - - - (1)
30x + 50y = 30000 - - - (2)
From (1)
x = 800 - y
Put x = 800 - y in (2)
30(800 - y) + 50y = 30000
24000 - 30y + 50y = 30000
24000 + 20y = 30000
20y = 30000 - 24000
20y = 6000
y =
Answer:
9g -5k
Step-by-step explanation: