Mathematics, 20.08.2019 00:30 sincere21

Use the elimination method
1) 3x+y=-1 5x-y=9
2) 4x+6y=24 4x-y=10
3)2x-y=-3 x+3y=16
4) 2x+3y=7 3x+4y=10

Answers: 1

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1) 3x+y=-1 5x-y=9
First of all we have add both equation but to be sure that the value we want to eliminate are both in a way that would make it possible t be deleted.
  3x+y=-1
  5x-y= 9
8x = 8
x= 1
In this case we are able to eliminate y becuase if we add +y-y we get that our answer is 0. and 3x + 5x would be 8x and -1+9 would be equal to 8 and to find x we needed to divided giving us that the answer for x is 1 becuase 8/8 is 1.
Then to find y we substitude the value of x in any of the formulas.
3(1)+y= -1
3+y= -1
y= -1-3
y=-4
When we have our y value we can determine if it is correct by replace the values.
5(1)--4= 9
5+4= 9
9=9
Up until now we are fine. So we do the same with the other equation.
3(1)+-4=-1
3+-4=-1
-1=-1
So by this we can now detemine that.
x= 1
y= -4

2) 4x+6y=24 4x-y=10
4x+ 6y =24
4x-y=10 (*-1)
4x+6y=24
-4x+y=-10
7y= 14
y= 14/7
y= 2
In this case we are not able to delete any of the variables so we multiplied by -1 to be able to eliminate x.
Then to find x we substitute the value of y in any of the formulas.
4x-2=10
4x= 10+2
x= 12/4
  x= 3
So we now know our variables so we substituted them to see if they are correct.
  4(3)+6(2)=24
12+12=24
24=24
We do the same with the other equation.
  4(3)-2=10
12-2 =10
   10= 10
So we can assume that.
x= 3
y= 2

3)2x-y=-3 x+3y=16
(3*)2x- y= -3
x+ 3y = 16
6x -3y = -9
x+3y =16
7x= 7
x= 1
In this case we are not able to delete any of the variables so we multiplied by 3 to be able to eliminate y.
Then to find y we substitute the value of x in any of the formulas.
1+ 3y = 16
3y= 16-1
y= 15/3
y= 5
So we now know our variables so we substituted them to see if they are correct.
2(1)- 5 =-3
2-5= -3
-3= -3
We do the same with the other equation.
1+3(5)= 16
1+15=16
16=16
So we now are sure that
x= 1
y= 5

4) 2x+3y=7 3x+4y=10
2x+3y =7 ( * - 4)
3x+4y =10 ( * 3)
-8x -12y = -28
9x +12y = 30
x= 2
In this case we are not able to delete any of the variables so we multiplied one of teh quations by - 4 to be able to subtract in our sum and the other by 3 to have the same number on y to be able to eliminate y.
Then to find y we substitute the value of x in any of the formulas.
2(2)+3y= 7
4+3y=7
3y= 7-4
y= 3/3
y= 1
So we now know our variables so we substituted them to see if they are correct.
3(2)+4(1)= 10
6+4=10
10=10
We do the same for the other
2(2)+3(1)=7
4+3= 7
7=7
So with that we can say that.
x= 2
y= 1

Answer from: Quest

If i’m not wrong i believe it is c

Answer from: Quest

mr. flanders’ class sold 47 doughnuts and mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

$\left\{\begin{array}{l}x+y=85\\1.35x+1.08y=104.49\end{array}\right.\rightarrow \left\{\begin{array}{l}x=85-y\\1.35(85-y)+1.08y=104.49\end{array}\right.$

$114.75-1.35y+1.08y=104.49,\\ \\114.75-104.49=1.35y-1.08y,\\ \\10.26=0.27y,\\ \\y=38,\\ \\x=85-38=47.$

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Use the elimination method
1) 3x+y=-1 5x-y=9
2) 4x+6y=24 4x-y=10
3)2x-y=-3 x+3y=16...

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