solve using the elimination method.show your work. if the system has no solution or infintite number of solutions, state this 4x-6y=30, 3x-5y=24
4x-6y=30
3x-5y=24
(4x-6y=30) Multiply with 3
12x-18y=90
(3x-5y=24) Multiply with 4
12x-20y=96
Now you have System:
12x-18y=90
12x-20y=96
Subtracting 12x-18y=90 from 12x-20y=96 gives:
12x-12x-18y-(-20y)=90-96
0-18y+20y= -6
2y= -6 Divide with 2
y= -3
4x-6y=30
4x-6*(-3)=30
4x+18=30
4x=30-18
4x=12 Divide with 4
x=3
OR
3x-5y=24
3x-5*(-3)=24
3x+15=24
3x=24-15
3x=9 Divide with 3
x=3
Solution is: x=3 y= -3
To solve this system of equations using the elimination method, we need to eliminate one of the variables by adding or subtracting the two equations.
Given the system of equations:
1) 4x - 6y = 30
2) 3x - 5y = 24
To eliminate a variable, we need to make the coefficients of either x or y the same in both equations. To do this, we can multiply the first equation by 3 and the second equation by 4:
Multiply equation 1 by 3:
3 * (4x - 6y) = 3 * 30
12x - 18y = 90
Multiply equation 2 by 4:
4 * (3x - 5y) = 4 * 24
12x - 20y = 96
Now, we have two equations with the same coefficient for x. We can subtract one equation from the other to eliminate x:
(12x - 18y) - (12x - 20y) = 90 - 96
12x - 18y - 12x + 20y = -6
-18y + 20y = -6
2y = -6
y = -6/2
y = -3
Substitute the value of y into one of the original equations to solve for x:
4x - 6(-3) = 30
4x + 18 = 30
4x = 30 - 18
4x = 12
x = 12/4
x = 3
So, the solution to the system of equations is x = 3 and y = -3.
Therefore, the system has a unique solution.