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Karo-lina-s [1.5K]
3 years ago
14

Solve for c 6/15=2/c

Mathematics
1 answer:
saul85 [17]3 years ago
7 0
\frac{6}{15} = \frac{2}{c}   Multiply both sides by 15
6 = \frac{30}{c}   Multiply both sides by c
6c = 30   Divide both sides by 6
c = 5   
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Help!!I don't know how to answer this which one is it ???
7nadin3 [17]

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5 0
3 years ago
Use the elimination method
Natalija [7]
<span>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
</span>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.<span>
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
</span>In this case we are not able to delete any of the variables so we multiplied by -1 to be able to eliminate x. <span>
Then to find x we substitute the value of y in any of the formulas.</span><span><span>
     </span>4x-2=10<span> 
     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
</span></span>In this case we are not able to delete any of the variables so we multiplied by 3 to be able to eliminate y. <span>
Then to find y we substitute the value of x in any of the formulas.</span>
<span>    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. <span>
Then to find y we substitute the value of x in any of the formulas.</span>
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</span>
5 0
3 years ago
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