4x - 6y = -48
-2x + 4y = 30
Solve the system of two equations using elimination with multiplication
To eliminate one of the variables, we need to multiply one or both of the equations by a constant so that the coefficients on one of the variables will cancel out when the two equations are added together.
Let's start with the first equation: 4x - 6y = -48
Let's multiply the second equation by 3 so that the coefficient on y will be opposite of the coefficient on y in the first equation:
-2(3x) + 4(3y) = 30(3)
-6x + 12y = 90
Now we have: -6x + 12y = 90
Adding this new equation to the first equation:
4x - 6y + (-6x + 12y) = -48 + 90
-2x + 6y = 42
Now we have one equation with only one variable:
-2x + 6y = 42
Now, we can solve for x in terms of y:
x = (6y - 42) / -2
x = -3y + 21
Now that we have found x in terms of y, we can substitute this expression into one of the original equations to solve for y:
-2(-3y + 21) + 4y = 30
6y - 42 + 4y = 30
10y = 72
y = 7.2
Now that we have found the value of y, we can substitute it back into our equation for x:
x = -3(7.2) + 21
x = -21.6 + 21
x = -0.6
Therefore, the solution to the system of equations is x = -0.6 and y = 7.2.
wrong its ether 3, 9, -3, 6, -21, -6, -33, -9
Let's go through the steps of solving the system of equations again:
Given equations:
4x - 6y = -48
-2x + 4y = 30
Step 1: Multiply the second equation by 2 to match the coefficient of x in the first equation:
-4x + 8y = 60
Step 2: Add the two equations together to eliminate x:
4x - 6y + (-4x + 8y) = -48 + 60
2y = 12
y = 6
Step 3: Substitute y = 6 into one of the original equations to solve for x:
-2x + 4(6) = 30
-2x + 24 = 30
-2x = 6
x = -3
Therefore, the correct solution to the system of equations is x = -3 and y = 6.
3x - 4y = 40
5x - 5y = 55
solve the system of equations
Given equations:
3x - 4y = 40
5x - 5y = 55
Step 1: Multiply the first equation by 5 and the second equation by 3 to make the coefficients of x equal:
15x - 20y = 200
15x - 15y = 165
Step 2: Subtract the second equation from the first equation to eliminate x:
(15x - 20y) - (15x - 15y) = 200 - 165
-5y = 35
y = -7
Step 3: Substitute y = -7 into one of the original equations to solve for x:
3x - 4(-7) = 40
3x + 28 = 40
3x = 12
x = 4
Therefore, the solution to the system of equations is x = 4 and y = -7.
how many solutions does the system have?
12x - 15y = 18
4x - 5y = 6
To determine the number of solutions for the system of equations, we can first rewrite the equations in the form Ax + By = C:
12x - 15y = 18
4x - 5y = 6
Now, we can check if the two equations are multiples of each other. If the equations are multiples of each other, they represent the same line and will have infinitely many solutions. Otherwise, if the equations are not multiples of each other, they will have a unique solution.
To check if the equations are multiples of each other, we can divide the coefficients of one equation by the coefficients of the other equation:
(12/4) = 3
(-15/-5) = 3
18/6 = 3
Since the coefficients of x and y in the two equations are equal (both are equal to 3), the two equations are multiples of each other and represent the same line. Therefore, the system of equations has infinitely many solutions.