To multiply the second equation to eliminate y, you would multiply the entire equation by 3.
1. 6x + 3y = 14
2. 9x + 6y = 30
Now you can subtract the first equation from the second equation to eliminate y:
(9x + 6y) - (6x + 3y) = 30 - 14
3x = 16
x = 16/3
Now substitute x back into the first equation to solve for y:
6(16/3) + 3y = 14
32 + 3y = 14
3y = -18
y = -6
Therefore, the solution to the system of equations using the elimination method is x = 16/3 and y = -6.
use the elimination method to solve this and what integer do you multiply in the 2nd equation
1. 6x + 3y = 14
2.3x + 2y = 10
5 answers
so, what Integer do you multiply the 2nd equation?
Apologies for the confusion in my previous response.
To eliminate y, we need to multiply the second equation by 3.
2. 3x + 2y = 10
2 * 3: 6x + 4y = 30
Now the system of equations is:
1. 6x + 3y = 14
2. 6x + 4y = 30
Now subtract the first equation from the second equation:
(6x + 4y) - (6x + 3y) = 30 - 14
y = 16
Now substitute y back into the first equation to solve for x:
6x + 3(16) = 14
6x + 48 = 14
6x = -34
x = -34/6
x = -17/3
Therefore, the solution to the system of equations using the elimination method is x = -17/3 and y = 16. And you should multiply the 2nd equation by 3 to eliminate y. Thanks for the clarification.
To eliminate y, we need to multiply the second equation by 3.
2. 3x + 2y = 10
2 * 3: 6x + 4y = 30
Now the system of equations is:
1. 6x + 3y = 14
2. 6x + 4y = 30
Now subtract the first equation from the second equation:
(6x + 4y) - (6x + 3y) = 30 - 14
y = 16
Now substitute y back into the first equation to solve for x:
6x + 3(16) = 14
6x + 48 = 14
6x = -34
x = -34/6
x = -17/3
Therefore, the solution to the system of equations using the elimination method is x = -17/3 and y = 16. And you should multiply the 2nd equation by 3 to eliminate y. Thanks for the clarification.
3 is not a option the options are 2, 6, -2 and -1/2
Apologies for the oversight. Let's reassess the options.
Given equations:
1. 6x + 3y = 14
2. 3x + 2y = 10
To eliminate y, we need to find a factor that when multiplied by the second equation will result in equal coefficients for y in both equations.
Looking at the options, if we multiply the second equation by -2, we get:
-2(3x + 2y) = -2(10)
-6x - 4y = -20
The system of equations becomes:
1. 6x + 3y = 14
2. -6x - 4y = -20
Now, add the two equations:
6x + 3y + (-6x) + (-4y) = 14 + (-20)
-y = -6
y = 6
Substitute y back into the first equation to find x:
6x + 3(6) = 14
6x + 18 = 14
6x = -4
x = -4/6
x = -2/3
Therefore, the solution to the system of equations when the second equation is multiplied by -2 is x = -2/3 and y = 6. The correct integer to multiply the second equation is -2.
Given equations:
1. 6x + 3y = 14
2. 3x + 2y = 10
To eliminate y, we need to find a factor that when multiplied by the second equation will result in equal coefficients for y in both equations.
Looking at the options, if we multiply the second equation by -2, we get:
-2(3x + 2y) = -2(10)
-6x - 4y = -20
The system of equations becomes:
1. 6x + 3y = 14
2. -6x - 4y = -20
Now, add the two equations:
6x + 3y + (-6x) + (-4y) = 14 + (-20)
-y = -6
y = 6
Substitute y back into the first equation to find x:
6x + 3(6) = 14
6x + 18 = 14
6x = -4
x = -4/6
x = -2/3
Therefore, the solution to the system of equations when the second equation is multiplied by -2 is x = -2/3 and y = 6. The correct integer to multiply the second equation is -2.
1 answer