Solve the system of two equations using elimination with multiplication.
4x − 6y = −48
−2x + 4y = 30
A. (−33,−9)
B. (3, 9)
C. (−3, 6)
D. (−21,−6)
First, we need to multiply the second equation by 2 to eliminate the x variable:
4x − 6y = −48
−4x + 8y = 60
Now, we can add the two equations together to eliminate the x variable:
(4x − 6y) + (−4x + 8y) = −48 + 60
2y = 12
y = 6
Now that we have y, we can substitute it back into one of the original equations to solve for x:
4x − 6(6) = −48
4x - 36 = -48
4x = -12
x = -3
Therefore, the solution to the system of equations is (-3, 6), which corresponds to option C.
Solve the system of equations.
3x − 4y = 40
5x − 5y = 55
A. (14, 3)
B. (4,−7)
C. (18, 7)
D. (8,−3)
To solve this system of equations using elimination with multiplication, we need to first create opposite coefficients for either the x or y variable in one of the equations. To do this, we will multiply both sides of the first equation by 5 and the second equation by 3:
First equation: 3x - 4y = 40
5(3x - 4y) = 5(40)
15x - 20y = 200
Second equation: 5x - 5y = 55
3(5x - 5y) = 3(55)
15x - 15y = 165
Now, we can subtract the second equation from the first equation to eliminate the x variable:
(15x - 20y) - (15x - 15y) = 200 - 165
-5y = 35
y = -7
Substitute y = -7 back into the first equation to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
Therefore, the solution to the system of equations is (4, -7), which corresponds to option B.
How many solutions does the system have?
12x − 15y = 18
4x − 5y = 6
A. one solution
B. infinitely many solutions
C. no solution
D. two solutions
To determine the number of solutions for this system of equations, we can check if the two equations are dependent (same line, infinitely many solutions) or inconsistent (parallel lines, no solution).
One way to do this is to divide one equation by the other and see if we get a constant ratio. Let's divide the first equation by the second equation:
(12x - 15y) / (4x - 5y) = 18 / 6
3 = 3
Since we get a constant value (3 = 3), this means the two equations are dependent, indicating that they represent the same line and there are infinitely many solutions.
Therefore, the system of equations has infinitely many solutions, which corresponds to option B.