Solve the system of equations using elimination: minus, 3, x, plus, 2, y, equals, 10−3x+2y=10 and 5, x, minus, 5, y, equals, minus, 105x−5y=−10
To solve the system of equations using elimination, we will eliminate one of the variables by adding or subtracting the equations. Let's eliminate the x variable.
First, let's multiply the second equation by 3 to make the x coefficients equal:
(3) × (5x - 5y = -10)
15x - 15y = -30
Now, we can add the new second equation to the first equation:
(10 - 3x + 2y) + (15x - 15y) = 10
Combine like terms:
10 + 15x - 3x + 2y - 15y = 10
Simplify:
12x - 13y = 0
Now, we have a new equation:
12x - 13y = 0 .............(3)
Let's use equations (1) and (3) to eliminate the x variable. Multiply equation (1) by 12:
12(10 - 3x + 2y) = 12(10)
120 - 36x + 24y = 120
Now, we can subtract equation (3) from the new equation:
(120 - 36x + 24y) - (12x - 13y) = 120
120 - 36x + 24y - 12x + 13y = 120
Combine like terms:
-48x + 37y = 0
Now, we have another new equation:
-48x + 37y = 0 .............(4)
Now, we have a system of equations:
12x - 13y = 0 .............(3)
-48x + 37y = 0 .............(4)
To eliminate the x variable, let's multiply equation (4) by 3:
3(-48x + 37y) = 3(0)
-144x + 111y = 0
Now, we can subtract equation (3) from equation (4):
(-144x + 111y) - (12x - 13y) = 0
-144x + 111y - 12x + 13y = 0
Combine like terms:
-156x + 124y = 0
Now, we have a new equation:
-156x + 124y = 0 .............(5)
Equation (5) is equivalent to equation (4).
Therefore, the system of equations is:
12x - 13y = 0 .............(3)
-156x + 124y = 0 .............(5)
To solve the system, we can set one of the variables to a constant, such as y = 1.
Let's solve equation (3) for x:
12x - 13y = 0
12x - 13(1) = 0
12x - 13 = 0
12x = 13
x = 13/12
So, the solution to the system of equations is x = 13/12 and y = 1.