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
solve each of them one by one
1.
Starting with the system of equations:
-3x - 4y = -23
3x - 5y = 5
To solve, we can add the two equations together to eliminate the x-terms:
-3x - 4y + 3x - 5y = -23 + 5
-y = -18
y = 18
Now, substitute y back into one of the original equations to solve for x:
3x - 5(18) = 5
3x - 90 = 5
3x = 95
x = 95/3
x = 31.67
Therefore, the solution to the system of equations is x = 31.67, y= 18.
2.
Since the two equations are identical when you ignore the signs, the system of equations has infinitely many solutions.
3.
Starting with the system of equations:
4x - 6y = -48
-2x + 4y = 30
To solve using elimination with multiplication, we can multiply the second equation by 2 to make the coefficients of x in both equations equal:
4x - 6y = -48
-4x + 8y = 60
Now, add the two equations together to eliminate x:
4y = 12
y = 3
Substitute y back into one of the equations to solve for x:
4x - 6(3) = -48
4x - 18 = -48
4x = -30
x = -30/4
x = -7.5
Therefore, the solution to the system of equations is x = -7.5, y= 3.
4.
Starting with the system of equations:
3x - 4y = 40
5x - 5y = 55
We can multiply the first equation by 5 and the second equation by -4 to create opposites for the x-terms:
15x - 20y = 200
-20x + 20y = -220
Now, add the two equations together to eliminate the y-terms:
-5x = -20
x = 4
Substitute x back into one of the equations to solve for y:
3(4) - 4y = 40
12 - 4y = 40
-4y = 28
y = -7
Therefore, the solution to the system of equations is x = 4, y= -7.
5.
Since the two equations are not multiples of each other, the system of equations has a unique solution.