1. Solve the system of two linear equations.
-3x - 4y = -23
3x - 5y = 5
2. How many solutions does the system of equations have?
8x - 6y = 48
8x + 6y = 48
3. Solve the system of two equations using elimination with multiplication.
4x - 6y = -48
-2x + 4y = 30
4. Solve the system of two equations.
3x - 4y = 40
5x - 5y = 55
5. How many solutions does the system have?
12x - 15y = 18
4x - 5y = 6
1. x = -2, y = 1
2. Infinite solutions
3. x = -6, y = 2
4. There is no solution
5. One solution
try again
1. x = 3, y = 4
2. Infinite solutions
3. x = -6, y = 6
4. x = 3, y = -1
5. One solution
you are wrong on 1, 3, and 4.
I apologize for the errors. Let me correct them:
1. x = 3, y = 4
3. x = -6, y = -3
4. There is no solution
Thank you for pointing out the mistakes.
3. Solve the system of two equations using elimination with multiplication.
4x - 6y = -48
-2x + 4y = 30
To use elimination with multiplication, we need to make the coefficients of either x or y in both equations the same (but opposite in sign). In this case, we can multiply the second equation by 3 to have oppositely signed coefficients for x in both equations.
Original equations:
1) 4x - 6y = -48
2) -2x + 4y = 30
Multiply equation 2 by 3:
3) -6x + 12y = 90
Now, add equations 1 and 3:
4x - 6y - 6x + 12y = -48 + 90
-2x + 6y = 42
Now, we have:
-2x + 6y = 42
-2x + 6y = 42
Since the equations are the same, there are infinite solutions to this system.